Ask question

Ask question

All categories

2 years ago

10

A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra

kes and accelerates uniformly to a stop just as he reaches a wall 50m away.
Required:
Sketch the a graph showing position v.s time and velocity v.s time.

1 answer:

PIT_PIT [208]2 years ago

4 0

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

You might be interested in

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0

2 years ago

A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

8 0

1 year ago

A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o

AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$

3 0

2 years ago

A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only

ANTONII [103]

Answer:

Technician A is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago