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sertanlavr [38]

4 years ago

9

an object is hanging by a string from your rearview mirror. While you are decelerating at a constant rate from 25 m/s to rest in

6.0 s (a) what angle does the string make with the vertical? (b) is it toward th windshiled or away from it

1 answer:

viktelen [127]4 years ago

5 0

This question has already previously been answered. :)

Here it is: brainly.com/question/2141424

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What is the kinetic energy of the rocket with mass 15,000 kg and speed of 5200 m/s? A. 2.01 x 10^11 J B. 2.02x 10^11 J C. 2.03 x

kow [346]

C. $E_{k}=2.03x10^{11}J$

The kinetic energy of a body is the ability to perform work due to its movement given by the equation $E_{k}=\frac{1}{2}mv^{2}$ .

To calculate the kinetic energy of a rocket with mass 15000kg and speed of 5200m/s:

$E_{k}=\frac{1}{2}(15000kg)(5200m/s)^{2}=202800000000J=2.03x10^{11}J$

8 0

4 years ago

Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of its displacem

Serga [27]

Answer:

True

Explanation:

Distance is defined as the length of the actual path traveled by the body.

Displacement is defined as the minimum distance between the two points.

the magnitude of displacement is always less than or equal to the distance traveled by the body.

As a deer runs from A to b , so it means the distance traveled by the deer is either equal to the magnitude of displacement or always greater than the magnitude of displacement of the deer.

Displacement can never be greater than the distance.

Thus, the option is true.

6 0

3 years ago

This is a net gain or loss of electrons. *

krok68 [10]

A net gain of electrons.

8 0

3 years ago

Read 2 more answers

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago

Ryan is playing with a ball. When would the ball<br> have the greatest polential energy?

forsale [732]

When the ball is not moving

7 0

4 years ago