During The Investigation Of A Traffic Accident, Police Find Skid Marks 55 Meters Long. They Determine (2024)

Given,

Two capacitors having values of 10 μF each are connected in parallel.

They are then connected to a 9.0 V battery.

The equivalent capacitance is:

[tex]C=10\mu F+10\mu F=20\mu F[/tex]

Thus the energy is equal to,

[tex]\begin{gathered} U=\frac{1}{2}CV^2 \\ \Rightarrow U=\frac{1}{2}\times(20\times10^{-6})\times9^2 \\ \Rightarrow U=8.1\times10^{-4}J \end{gathered}[/tex]

The answer is

[tex]8.1\times10^{-4}J[/tex]

Answer:

The weight of the vehicle on Mars = 2191.95 N

Explanation:

Note that:

Mass is the quantity of matter in an object

The weight is the product of mass and the acceleration due to gravity

The acceleration due to gravity on Mars = 0.38g

where g = 9.81 m/s²

Mass of the vehicle = 588 kg

Weight = Mass x Acceleration due to gravity

Weight = 588 x 0.38g

Weight = 588 x 0.38 x 9.81

Weight = 2191.95 N

The weight of the vehicle on Mars = 2191.95 N

ANSWER:

F = m x a

STEP-BY-STEP EXPLANATION:

In the 3 exposed systems, we have in common that Newton's 2nd law applies, which establishes that if a net force is applied to an object, the speed of the object will change given that its direction or speed will change (acceleration), that the force varies according to the acceleration exerted on an object with stated mass.

Therefore, the common equation is:

[tex]F=m\cdot a[/tex]

14Y-2 = 2Y + 140

[tex]\begin{gathered} 14y-2y=140+2 \\ 12y=142 \\ y=\frac{142}{12} \\ =11.83 \end{gathered}[/tex]

Thus, the value of unknown y is 11.83.

The reaction of the nail is obtained as 26.5 N towards the wall.

What is the magnitude of the reaction?

We know that there is a resultant force that the nail would exert on the wall. The wall would then exert back a reaction force on the nail. Thus we ought to obtain that force as well as its direction and we would do that by the use of the parallelogram law of vectors.

We have;

R^2 = a^2 + b^2 - 2abCos θ

Where a and b refers to the forces 10N and 20N respectively and the angle between them is 120 degrees.

R^2 = 10^2 + 20^2 - 2(10 * 20)Cos 120

R^2 = 100 + 400 - 400Cos120

R^2 = 500 + 200

R^2 = 700

R = 26.5 N

Learn more about reaction force:https://brainly.com/question/14853868

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ANSWER:

36.24 m/s

STEP-BY-STEP EXPLANATION:

Given:

Mass (m) = 54 kg

Height (h) = 67 m

From the law of conservation of energy, we have that the initial energy is equal to the final energy

The initial energy is just gravitational potential energy since the skier is not moving at the bottom of the hill and the final energy is the potential energy that is converted into kinetic and thermal energy due to friction, if we ignore friction, we would be left with the following:

[tex]\begin{gathered} E_P=E_C \\ \\ mgh=\frac{1}{2}mv^2 \\ \\ gh=\frac{1}{2}v^2 \\ \\ \text{ We replacing} \\ \\ (9.8)(67)=\frac{1}{2}v^2 \\ \\ v^2=2\cdot656.6 \\ \\ v=\sqrt{1313.2} \\ \\ v=36.24\text{ m/s} \end{gathered}[/tex]

The speed of the skier on reaching the bottom of the slope is 36.24 m/s

Answer:

= : Displacement from t = 0 s to t = 5s

A: Displacement from t = 0 s to t = 2.5s

B: Displacement from t = 2.5 s to t = 5s

= : Average velocity from t = 0 s to t = 5s

A : Average velocity from t = 0 s to t = 2.5s

B: Instantaneous velocity at t = 5s

Explanation:

The displacement is the difference between the final and the initial position.

Since at t = 0 and t = 5, A and B have the same position, the displacement was equal. So, the first answer is

= : Displacement from t = 0 s to t = 5s

In the same way, we can see that at t = 2.5 the position of A is greater than the position of B, so from t = 0 to t = 2.5, A has a greater difference of position but from t = 2.5 to t = 5, B has a greater difference of position. Then:

A: Displacement from t = 0 s to t = 2.5s

B: Displacement from t = 2.5 s to t = 5s

The average velocity is equal to the displacement divided by the time. if the time is equal, the average velocity will be greater if the displacement is greater. So:

= : Average velocity from t = 0 s to t = 5s

A : Average velocity from t = 0 s to t = 2.5s

Finally, the instantaneous velocity is the slope of the graph at t = 5. The greater the slope, the greater the velocity. Since the slope of B is greater than the slope of A at t = 5, we can say that B has a greater instantaneous velocity at t =5, so:

B: Instantaneous velocity at t = 5s

So, the answers are:

= : Displacement from t = 0 s to t = 5s

A: Displacement from t = 0 s to t = 2.5s

B: Displacement from t = 2.5 s to t = 5s

= : Average velocity from t = 0 s to t = 5s

A : Average velocity from t = 0 s to t = 2.5s

B: Instantaneous velocity at t = 5s

Given,

The radius of the cylinder and the hemisphere, r=3 inches

The height of the cylinder, h=10 inches.

The volume of the whole figure is the sum of the volume of the cylinder and the volume of the hemisphere.

Hemisphere is the half of a sphere. Thus the volume of a hemisphere is equal to half of the volume of the sphere with the same radius as the hemisphere.

The volume of the given hemisphere is,

[tex]V_h=\frac{2}{3}\pi r^3[/tex]

On substituting the known values,

[tex]\begin{gathered} V_h=\frac{2}{3}\pi\times3^3 \\ =56.55\text{ cubic inches} \end{gathered}[/tex]

The volume of the cylinder is given by,

[tex]V_c=\pi r^2h[/tex]

On substituting the known values,

[tex]\begin{gathered} V_c=\pi\times3^2\times10 \\ =282.74\text{ cubic inches} \end{gathered}[/tex]

Thus the total volume of the given figure is,

[tex]\begin{gathered} V=V_h+V_c \\ =56.55+282.74 \\ =339.29\text{ cubic inches} \end{gathered}[/tex]

Thus the volume of the given figure is 339.29 cubic inches

We will have the following:

*If a person has a weigth of 500N on Earth, and moves from the ground level to a position high up on a mountain this will affect the gravitational force between the person and the Earth in a very minuscule way. The net change is given only in the gravitational potential energy stored in the person due to the position higher up- And the force between the person and the Earth will be slighly smaller, since the distance is a little bit greater between the center of the Earth and the person.

*We have that they are both acceleration towards each other, the person is falling towards the Earth and the Earth is falling towards the person, but the acceleration of the person towards the Earth is orders of magnitude greater than the Earth falling to the person.

Given data

*The given mass of the rocket is m = 14000 kg

*The given momentum is p = 2.8 × 10^7 kg.m/s

The formula for the velocity of the rocket is given by the momentum formula as

[tex]\begin{gathered} p=mv \\ v=\frac{p}{m} \end{gathered}[/tex]

Substitute the known values in the above expression as

[tex]\begin{gathered} v=\frac{2.8\times10^7}{14000} \\ v=2000\text{ m/s} \end{gathered}[/tex]

Hence, the velocity of the rocket is v = 2000 m/s

[tex]\begin{gathered} v=5\text{ m/s} \\ t=2\text{ s} \\ F=140N \\ m=\text{?} \\ F=\text{ ma} \\ \text{Solving m} \\ m=\frac{F}{a} \\ \text{but} \\ a=\frac{v}{t} \\ a=\frac{5\text{ m/s}}{2s} \\ a=2.5m/s^2 \\ \text{Hence} \\ m=\frac{140N}{2.5m/s^2} \\ m=56\operatorname{kg} \\ \text{The mass is 56 kg} \end{gathered}[/tex]

We know that the acceleration is defined as:

[tex]a=\frac{dv}{dt}[/tex]

This means that if we know the acceleration we can find the velocity with the integral:

[tex]v=\int ^t_{t_0}adt^{\prime}+v_0[/tex]

where t0 denotes some intial time and v0 the velocity at that time.

In this case we know that the acceleration is 4m/s^2 and that at time t=1 the velocity is 4 m/s, then we have:

[tex]\begin{gathered} v=\int ^t_14dt^{\prime}+4 \\ =4(t^{\prime})\vert^t_1+4 \\ =4(t-1)+4 \\ =4t-4+4 \\ =4t \end{gathered}[/tex]

hence the velocity function for this motion is:

[tex]v=4t[/tex]

Now, we know that the velocity is defined by:

[tex]v=\frac{dx}{dt}[/tex]

then the position can be obtained by:

[tex]x=\int ^t_{t_0}vdt^{\prime}+x_0[/tex]

where t0 is some time and x0 is the position at that time; in this case we don't have an initial position for the particle then te position will be given by:

[tex]\begin{gathered} x=\int ^t_{t_0}4t^{\prime}dt^{\prime}+x_0 \\ x=4(\frac{t^{\prime2}}{2})\vert^t_{t0}+x_0 \\ x=4(\frac{t^2}{2}-\frac{t^2_0}{2})+x_0 \\ x=2t^2-2t^2_0+x_0 \end{gathered}[/tex]

Hence the position at any given time is given by:

[tex]x=2t^2-2t^2_0+x_0[/tex]

Once we know the position function we can calculate the area undert the position time graph in the interval given:

[tex]\begin{gathered} \int ^3_1(2t^2-2t^2_0+x_0)dt=(\frac{2}{3}t^3+(x_0-2t^2_0)t)\vert^3_1 \\ =\frac{2}{3}(3^3-1^3)+(x_0-2t^2_0)(3^{}-1^{}) \\ =\frac{2}{3}(27-1)+(x_0-2t^2_0)(2) \\ =\frac{2}{3}(26)+(x_0-2t^2_0)(2) \\ =\frac{52}{3}+2(x_0-2t^2_0) \end{gathered}[/tex]

Therefore, in general, the area under the curve in the given interval is:

[tex]\frac{52}{3}+2(x_0-2t^2_0)[/tex]

If we assume that the particle was at the origin at time t0=0 (this meas x0=0 as well), then the area under the position-time graph will be:

[tex]\frac{52}{3}+2(0-2(0)^2)=\frac{52}{3}[/tex]

Note: No matter what the intial time and position is the expression we found will give us the correct answer.

the frequency is 30Hz

[tex]\begin{gathered} W=70J \\ F=15N \\ d=\text{?} \\ W=Fd \\ \text{Solving d } \\ d=\frac{W}{F}=\frac{70J}{15N}=4.67m \\ d=4.67\text{ m} \\ The\text{ distance He moved the box is 4.67m} \\ \\ \text{Question 2)} \\ False,\text{ in p}\operatorname{erf}ect\text{ conditions the efficiency will be 100\% not greater } \end{gathered}[/tex]

We will have the following:

We remember that the distance for a linear accelerated motion is given by the expression:

[tex]d=v_it+\frac{1}{2}at^2[/tex]

Here "d" is the distance, "vi" is the initial velocity (In our case it starts at rest so 0 m/s), "t" is the time and "a" is the acceleration at which the object is subjected, so:

[tex]d=(0m/s)(5.00s)+\frac{1}{2}(10.0m/s^2)(5.00s)^2\Rightarrow d=125.0m[/tex]

So, the distance the car travels in those 5 seconds is 125.0 m.

Potential energy (PE) = m g h

Where:

m = mass = 425 kg

g = gravity = 9.8 m/s2

h = height

PE = 72250J

Replacing:

72250 J = 425 kg * 9.8 m/s^2 * h

Solve for H

72250 J/ (425 kg * 9.8 m/s^2) = h

h = 17.35 m

Given:

Mass of car, m = 2000 kg

Coefficient of kinetic friction, μk = 0.70

Distance = 75 m

Let's find the speed of the car.

To find the speed of the car, apply the motion formula:

[tex]v^2=u^2+2as[/tex]

Where:

v is the final velocity = 0 m/s

u is the initial velocity

a is the acceleration of the car.

s is the distance = 75 m

To find the acceleration, we have:

[tex]a=-\mu_k*g[/tex]

Where:

μk = 0.70

g is acceleration due to gravity = 9.8 m/s²

Thus, we have:

[tex]\begin{gathered} a=-0.70*9.8 \\ \\ a=-6.86\text{ m/s}^2 \end{gathered}[/tex]

The deceleration of the car is -6.86 m/s².

Now, to find the initial velocity u, we have:

[tex]\begin{gathered} v^2=u^2+2as \\ \\ 0^2=u^2+2(-6.86)(75) \\ \\ 0=u^2-1029 \\ \\ u^2=1029 \end{gathered}[/tex]

Take the square root of both sides:

[tex]\begin{gathered} \sqrt{u^2}=\sqrt{1029} \\ \\ u=32.08\text{ m/s} \end{gathered}[/tex]

Therefore, the speed of the car was 32.08 m/s.

ANSWER:

32.08 m/s

Answer:

See below

Explanation:

There is no friction so we will not need to calculate the Fn

we only need the Fdp = Force downplane = mgsin 36 = 69.19 N

F = ma

69.19 = 12 a a = 5.766 m/s^2

df = 3 = do + vo t + 1/2 at^2

3 = 0 + 0 + 1/2 (5.766) t^2 t = 1.02 s

vf = vo + at

vf = 0 + 5.766 ( 1.02) = 5.88 m/s

KE when it hits the spring = 1/2 (12)(5.88)^2= 207.58 J

The work of compressing the spring must equal this

Ws = 1/2 kx^2 207.58 = 1/2 (1 x10^4) x^2 shows x = .204 m

(Check my math !)

D. if I increases, R must decrease

Explanation

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.it says

the triangle states the 3 formulas:

[tex]\begin{gathered} V=I*R \\ I=\frac{V}{R} \\ R=\frac{V}{I} \end{gathered}[/tex]

so

Step 1

if we let V constant

[tex]\begin{gathered} V=R*I \\ constant(k)=R*I \end{gathered}[/tex]

now, as the product of R and I must be constant:

if you increase the resistance, the current must decrease, or if yuo increase the current, the resistance must decrease to maintain the same voltage ( constant)

therefore, the correct answer is

D. if I increases, R must decrease

I hope this helps you

33 1/3 rpm can be written as follow;

[tex]33\frac{1}{3}\text{rpm}=\frac{100}{3}rpm[/tex]

The period is given by:

[tex]T=\frac{2\pi}{\omega}[/tex]

where w = 100/3 rpm is the angular frequency. Convert 100/3 rpm to rev/s, as follow:

[tex]\frac{100}{3}\frac{\text{rev}}{\min}\cdot\frac{1\min }{60s}=0.555\frac{rev}{s}[/tex]

By replacing into the formula for the period T you obtain:

[tex]T=\frac{2\pi}{0.555\frac{\text{rev}}{s}}\approx11.3s[/tex]

Hence, the period is approximately 11.3s

ANSWER:

1.3 times faster

STEP-BY-STEP EXPLANATION:

Given:

Distance ostrich = 56 m

Time ostrich = 14.5 s

Initial velocity = 0 m/s

To calculate the final velocity of the ostrich we use the following formula:

[tex]\begin{gathered} d=0.5(v_f+v_i)\Delta t \\ \\ \text{ we solve for }v_f \\ \\ 56=0.5(0+v_f)14.5 \\ \\ 56=7.25v_f \\ \\ v_f=\frac{56}{7.25}=7.7\text{ m/s} \end{gathered}[/tex]

Now, we calculate the average velocity of the ostrich, like this:

[tex]\begin{gathered} v_o=\frac{v_i+v_f}{2}=\frac{0+7.7}{2} \\ \\ v_o=3.9\text{ m/s} \end{gathered}[/tex]

We calculate the time it would take the ostrich to travel 100 meters with the following formula:

[tex]\begin{gathered} d=0.5(v_{f}+v_{i})\Delta t \\ \\ \text{ we replacing} \\ \\ 100=0.5\left(0+7.7\right)\Delta t \\ \\ \Delta t=25.9\text{ sec} \end{gathered}[/tex]

now we calculate the time of the cheetah knowing that it takes 5.7 seconds less:

[tex]\begin{gathered} t_c=t_o-5.7 \\ \\ t_c=25.9-5.7=20.2\text{ sec} \end{gathered}[/tex]

Now, we can calculate the final velocty of the cheetah using the following formula:

[tex]\begin{gathered} d=0.5(v_{f}+v_{i})\Delta t \\ \\ \text{ we replacing} \\ \\ 100=0.5\left(v_f+0\right)20.2 \\ \\ v_f=\frac{100}{10.1} \\ \\ v_f=9.9\text{ m/s} \end{gathered}[/tex]

Now we calculate the average velocity, like this:

[tex]\begin{gathered} v_c=\frac{v_i+v_f}{2}=\frac{0+9.9}{2} \\ \\ v_c=4.95\text{ m/s} \end{gathered}[/tex]

we calculate the ratio between the two to know how many times the cheetah is faster:

[tex]r=\frac{4.95}{3.9}=1.3[/tex]

Therefore, the cheetah is 1.3 times faster than the ostrich.

Given,

The mass of the ball, m=16 g=16×10⁻³ kg

The length of the string, r=1.4 m

The period of revolution of the ball, T=1.09 s

The angular velocity of the ball is given by,

[tex]\omega=\frac{2\pi}{T}[/tex]

On substituting the known values,

[tex]\begin{gathered} \omega=\frac{2\pi}{1.09} \\ =5.76\text{ rad/s} \end{gathered}[/tex]

The tension on the string will be equal to the centrifugal force that acts on the ball.

And it is given by,

[tex]F=m\omega^2r[/tex]

On substituting the known values,

[tex]\begin{gathered} F=16\times10^{-3}\times5.76^2\times1.4 \\ =0.74\text{ N} \end{gathered}[/tex]

Thus the magnitude of the tension of the string is 0.74 m

We are asked to determine the magnification of a concave mirror. To do that we will use the following formula:

[tex]M=\frac{h_i}{h_0}=-\frac{d_i}{d__0}[/tex]

Where:

[tex]\begin{gathered} M=\text{ magnification} \\ h_0=\text{ height of the object} \\ h_i=\text{ height of the image} \\ d_0=\text{ distance of the object } \\ di=\text{ distance of the image} \end{gathered}[/tex]

We are given the following values:

[tex]\begin{gathered} h_0=5cm \\ d_0=2.7cm \\ h_i=14.3cm \end{gathered}[/tex]

Therefore, we can use the following to determine the magnification:

[tex]M=\frac{h_i}{h_0}[/tex]

Substituting the values:

[tex]M=\frac{14.3cm}{5cm}[/tex]

Solving the operation:

[tex]M=2.86[/tex]

Therefore, the magnification is 2.86

acceleration = ( final velocity- initial velocity ) / time

Initial velocity = 30 m/s

Final velocity = 12 m/s

time = 8 seconds

Replacing:

a = (12 - 30) / 8 = -18/8 = -2.25 m/s2

Given:

The train first travels 450 km towards the east from Toronto, then 200 km towards the south, and finally 150 km [W 55 degrees S].

The time traveled is t = 12 hours.

To find The train's total displacement relative to Toronto.

The train's average velocity.

Explanation:

The vector diagram can be drawn as shown below

Let Toronto be at the origin.

Here, X-axis is in the East direction.

Y-Axis is in the North direction.

The displacement in the X-direction will be

[tex]\begin{gathered} d_x=450-150cos55^{\circ}\text{ } \\ =363.96km \end{gathered}[/tex]

The displacement in the Y-direction is

[tex]\begin{gathered} d_y=200+150\sin (55^{\circ})\text{ } \\ =\text{ 322.87 }km \end{gathered}[/tex]

The magnitude of resultant displacement will be

[tex]\begin{gathered} d=\sqrt[]{(d_x)^2+(d_y)^2} \\ =486.53km \end{gathered}[/tex]

The direction of displacement will be

[tex]\begin{gathered} \tan \theta=\frac{d_y}{d_x} \\ \theta=\tan ^{-1}(\frac{d_y}{d_x}) \\ =\tan ^{-1}(\frac{322.87}{363.96}) \\ =41.57\text{ degr}ees \end{gathered}[/tex]

(b) The train's average velocity will be

[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{486.53}{12} \\ =\frac{40.54km}{h} \end{gathered}[/tex]

Final Answer: (a) The magnitude of resultant displacement is 486.53 km.

The direction of resultant displacement is E 41.57 degrees S.

(b) The train's average velocity is 40.54 km/h.

In order to find the current needed, we can use the formula below:

[tex]I=\frac{V}{R}[/tex]

Where I is the current (in Ampere), V is the voltage (in Volts) and R is the resistance (in ohms)

So, using V = 120 V and R = 958 ohms, we have:

[tex]\begin{gathered} I=\frac{120}{958}\\ \\ I=0.1253\text{ A} \end{gathered}[/tex]

Therefore the current needed is 0.1253 Amperes.

Given:

• Vertical drop = 95 meters

,

• Initial velocity = 0 m/s

,

• Mass of cart and riders = 1000 kg

,

• Final velocity = 35 m/s

Let's find the amount of potential energy converted to thermal energy.

For the potential energy, we have:

PE = mgh

Also thermal energy is a form of kinetic energy.

Thus, for the thermal energy, we have:

[tex]KE=\frac{1}{2}mv^2[/tex]

Now, to find the difference, we have:

[tex]\begin{gathered} PE-KE \\ \\ =(\text{mgh)}-(\frac{1}{2}mv^2) \end{gathered}[/tex]

Where:

g is acceleration due to gravity = 9.8 m/s^2

v is the final velocity = 35 m/s

h is the height = 95 meters

m is the mass

Thus, we have:

[tex]\begin{gathered} (1000\times9.8\times95)-(\frac{1}{2}\times1000\times35^2) \\ \\ =931000-612500 \\ \\ =318500\text{ J }\approx319000\text{ }\approx319\text{ kJ} \end{gathered}[/tex]

Therefore, the amount of potential energy converted into thermal energy is 319 kiloJoules.

ANSWER:

319 kJ

ANSWER:

224.32 m/s

STEP-BY-STEP EXPLANATION:

We can use Bernoilli's theorem:

[tex]P_1+\frac{1}{2}\rho(V_1)^2=P_2+\frac{1}{2}\rho(V_2)^2[/tex]

We replace and solve for V2, like this:

[tex]\begin{gathered} P_1-P_2=\frac{1}{2}\rho(V_2)^2-\frac{1}{2}\rho(V_1)^2 \\ 9460=\frac{1}{2}\cdot0.376\cdot(V_2)^2-\frac{1}{2}\cdot0.376\cdot(0)^2 \\ 9460=\frac{1}{2}\cdot0.376\cdot(V_2)^2 \\ (V_2)^2=50319 \\ V_2=\sqrt[]{50319}=224.32\text{ m/s} \end{gathered}[/tex]

The airplane’s speed relative to the air is 224.32 m/s

ANSWER and EXPLANATION

We want to draw the free body diagram of the block being held against the wall.

There are four forces acting on the block:

1. Force from the push

2. Normal force

3. Gravitational force/Weight

4. Frictional force

The free-body diagram is shown below:

where m = mass of the block

Fr = Frictional force

N = Normal force

Fp = Force from the push

Fg = Gravitational force/Weight

Since the block is in equilibrium, the net force in both horizontal and vertical directions must be 0.

Therefore, we have:

[tex]\begin{gathered} F_{\text{hor}}=N-F_p=0 \\ \text{F}_{\text{ver}}=F_r-F_g=0 \end{gathered}[/tex]

As we can see, the magnitude of the Normal force will be equal to the magnitude of the force from the push and the magnitude of the Frictional force will be equal to the magnitude of the Gravitational force.

[tex]\begin{gathered} N=F_p \\ F_r=F_g \end{gathered}[/tex]

We cannot calculate some of the values because the mass of the block and the coefficient of static friction are not given, but we have that:

[tex]\begin{gathered} F_p=10N \\ \Rightarrow N=F_p=10N \end{gathered}[/tex]