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Amanda [17]
3 years ago
5

HELP ASAP WHAT ARE PROPERTIES OF WATER

Physics
1 answer:
iren [92.7K]3 years ago
3 0
Pls mark as brainiest!!

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What is the velocity of the car, in meters per second, just after it hits a 135-kg deer initially running at 10.5 m/s in the sam
Gnom [1K]

Answer:

V = 27.46 m/s

Explanation:

given,

mass of the deer(m) = 135 Kg

speed of the deer (u) = 10.5 m/s

assuming,

mass of the car(M) = 900 Kg

initial velocity of car (v) = 30 m/s

using conservation of momentum

m u + M v = (M + m )V

V is the velocity of the car as deer is on the car

135 x 10.5 + 900 x 30 = (900 + 135 ) V

28417.5 = 1035 V

V = 27.46 m/s

so, the velocity of car is equal to V = 27.46 m/s

5 0
4 years ago
Who has the greater acceleration rate? A runner
Arturiano [62]
The first runner because it is very clear that accelarition depends on the time and we know that the time in this case is pretty simple
6 0
4 years ago
If a football player has more mass they will also have more ______ ?<br> Fill in the blank
serg [7]

If a football player has more mass, they will also have more <u>momentum</u>. This is because mass is directly proportional to momentum.

3 0
3 years ago
Mass of a car is given as 6000 kg is moving with a velocity of 40 m/s. Calculate the momentum of the car
Alecsey [184]

Answer:

Explanation:

given

m = 6000kg

v = 40 m/

p = ?

solution

p = mv

p = 6000kg × 40m/

p =240000kg.m/

7 0
1 year ago
Read 2 more answers
Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera
spin [16.1K]

Answer:

R' = \frac{1}{\sqrt{3}}R

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

g' = 3g

So we would have

\frac{GM}{R'^2}=3(\frac{GM}{R^2})

Solving the equation for R', we find

R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R

7 0
3 years ago
Read 2 more answers
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