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2 years ago

9

A car driver sees a rabbit on the road. The driver makes an emergency stop after he sees the rabbit. Figure given shows the spee

d of the car from the time the driver sees the rabbit until the car stops. ) Calculate the distance that the car travels in the first 0.5 seconds.
2. A car of mass 1000 kg is moving with a velocity of 10 m/s. If the velocity-time graph for this car is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be:

2 answers:

svetoff [14.1K]2 years ago

4 0

The car must be very fast kuh on bro

Fed [463]2 years ago

3 0

It was nothing becauseuibdadvbipduadbcuipdbaduicphdsa

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, $mg < ks_s$

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

$P = mv$

$P = 0$

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

$F_{spring} > mg$

$Ks > mg$

6 0

3 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago

A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is:

$\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}$

$\alpha = 36.469\,\frac{rad}{s^{2}}$

$I = \frac{\tau}{\alpha}$

$I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }$

$I = 0.0823\,N\cdot m\cdot s^{2}$

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0

2 years ago

Which statement about activation energy is true? A. It is not affected by catalysts. B. It has no effect on the rate of the reac

HACTEHA [7]

C it is the energy required to break existing chemical bonds, it is the amount of energy that a reaction requires in order for the reactants to successfully collide and react

6 0

3 years ago

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