Ask question

Ask question

All categories

Natali5045456 [20]

3 years ago

8

A group of tourists drives through a state park in Maine. While arguing about whether the animal they've just seen is an Elk or

a Moose, a chipmunk darts across the park road. The driver, who was traveling 6.7 m/s through the park slams on their breaks and comes to a stop. If the car has an acceleration of -8.3 m/s2, how long does it take the car to stop?

1 answer:

podryga [215]3 years ago

6 0

Answer:

0.8 second

Explanation:

Given data

Speed v=6.7m/s

Deceleration = - 8.3m/s²

Initial velocity u= 0m/s

We can solve for the distance covered upon breaking the speed using the expression

v=u +at

6.7=0-(8.3)t

6.7=8.3t

Divide both sides by 8.3

6.7/8.3=t

t=6.7/8.3

t=0.8 second

You might be interested in

Which factor is least likely to limit the rate of photosynthesis?

madreJ [45]

Answer:

oxygen concentration

Explanation:

8 0

3 years ago

If it takes 100 N to get a 10 kg object to accelerate at 10m/s/s, how much force will it take to get a 20 kg object to accelerat

yan [13]

Answer: C) 200 N

Explanation:

The force $F$ is defined as:

$F=m.a$

Where:

$m=20 kg$ is the mass of the object

$a=10 m/s^{2}$ is the acceleration

Then:

$F=(20 kg)(10 m/s^{2})$

Finally:

$F=200 N$

Hence, the correct option is C.

7 0

3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra

Tatiana [17]

Answer:

The centripetal acceleration of the car is $8\ m/s^2$ .

Explanation:

Let the mass of the car, $m=10^3\ kg$

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(20\ m/s)^2}{50\ m}$

$a=8\ m/s^2$

So, the centripetal acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago