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faltersainse [42]
3 years ago
11

What force pulls all objects with mass toward one another?

Physics
1 answer:
Vilka [71]3 years ago
5 0
Gravity

Hope this helps:)
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zhannawk [14.2K]
Boyles law describes pressure volume relationship at constant temperature  pressure is directly proportional  to volume at constant tamparature
4 0
3 years ago
Read 2 more answers
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.
7 0
3 years ago
An object is thrown into the air at 60m/s, straight up. What is its velocity at the highest point?
Cloud [144]

Explanation:

Whenever an object is at its highest point, the velocity and acceleration of the object is zero.

6 0
3 years ago
A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl
guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a  = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b  = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

P_1 = m_av_a + m_b v_b

P_1 = 1300 \times 35+ 1000 \times 25

P_1 =70500 Kg.m/s

final momentum

P_2 = m_aV_a + m_b V_b

P_2 = 1300 \times 30+ 1000 \times 31.5

P_2 =70500 Kg.m/s

here  initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0
3 years ago
It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Ass
Vinvika [58]

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

<h3>Acceleration of the car </h3>

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

<h3>Distance traveled when the speed is 32 m/s</h3>

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

Learn more about distance here: brainly.com/question/4931057

#SPJ1

5 0
2 years ago
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