Ask question

Ask question

All categories

4 years ago

11

PLEASE HELP WILL NAME BRAINLIEST!! An object is dropped from rest. What is its instantaneous speed when it has been in motion fo

r 4 s? The acceleration of gravity is 9.8 m/s^2 Answer in units of m/s.

1 answer:

frosja888 [35]4 years ago

3 0

Explanation:

Using kinematics,

t = 4s

u = 0m/s

a = 9.8m/s^2

Therefore v = u + at = 0 + (4)(9.8) = 39.2m/s.

You might be interested in

Brody is barbecuing hamburgers on the grill. He notices that the coals are ready for cooking because he feels the warmth of the

hodyreva [135]

Answer:

i think the answer is infrared

8 0

3 years ago

A 1500 kg car is approaching the hill shown in the figure below at 10 m/s when it suddenly runs out of gas.

kenny6666 [7]

Based on the data, the car would reach a final velocity of 0 m/s. With a mass of 1500 kg, the momentum would be -15000 kg.m/sThank you for your question. Please don't hesitate to ask in Brainly your queries.

6 0

3 years ago

A circular coil that has =130 turns and a radius of =11.5 cm lies in a magnetic field that has a magnitude of 0=0.0725 T directe

Levart [38]

I think is the half of 130

7 0

3 years ago

a .6 kg ball has a speed of 2 m/s at point a and kinetic energy of 7.5 j at point b. determine the following: the balls speed

Vladimir79 [104]

Answer:

Can You not Answer yourself

3 0

3 years ago

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