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3 years ago

8

A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t

he mass of the car?

1 answer:

vodka [1.7K]3 years ago

4 0

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be $m$ kg.

Given:

Initial velocity of the car is, $u=25\ m/s$

As the car stops, final velocity of the car is, $v=0\ m/s$

Change in momentum is, $\Delta p=39400$

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

$\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg$

Therefore, the mass of the car is 1576 kg.

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Good morning! Can someone please answer this, ill give you brainliest and you will earn 50 points.

iren2701 [21]

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5 0

2 years ago

Read 2 more answers

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

3)

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The

n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

Explanation:

Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

v = √17. 023 × 10^34-27

v = √17.023 × 10^7

v = 13047.221

v = 1.31 × 10^4

4 0

4 years ago

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago