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fgiga [73]
3 years ago
14

What impulse results if the force acting on an object is described by the force-time graph shown?

Physics
1 answer:
Hoochie [10]3 years ago
8 0
The correct answer to this question is "2.1 kgm/s." The impulse results if the force acting on an object that is described by the force-time graph shown is that 2.1 kgm/s. Based on the graph, it shows like a plateau with the flat at the force of 2.1 N.
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1
Darina [25.2K]

Answer:

7.5 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1  = 60kg

Mass 2  = 75kg

Distance between the bodies  = 2m

Unknown:

Gravitational fore  = ?

Solution:

The gravitational force between the two bodies can be derived using;

  F  = \frac{G mass 1 x mass 2}{distance^{2} }  

    G is the universal gravitation constant  = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²

Insert the parameters and solve;

  F  = \frac{6.67 x 10^{-11}  x 60 x 75}{2^{2} }   = 7.5 x 10⁻⁸N

3 0
3 years ago
A car is moving with a uniform speed of 15.0 m/s along a straight path. What is the distance covered by the car in 12.0 minutes?
Ksivusya [100]
<span>1.08 x 10 ^1 km Hope i helped</span>
5 0
4 years ago
Read 2 more answers
Can anyone explain<br>if knows​
Aleks [24]

Answer:

Sry I’m just trying to get my points :(

Explanation:

Better luck nest time I would help if I was smart enough but currently I’m as d u m b as a rock...

5 0
3 years ago
Read 2 more answers
An object has a mass of 0.250 kg. What is the gravitational force of on the object by the earth?
sesenic [268]

Answer:

2.4525 N

Explanation:

The earths gravity is 9.81 N/Kg

And so to work this out you would multiply 9.81 by 0.250 which equals to 2.4525N

7 0
4 years ago
a boat is moving 1.3m/s N while the steam current is pushing the boat 2.2m/s What is the resultant velocity( I also need the giv
lubasha [3.4K]

Answer:

6) 2.6 m/s, 31°

7) 9.2 m/s

8) 1.2 s

Explanation:

I'll do #6, #7, and #8 as examples.  You can solve #9 using the equation from #7, and #10 using the equation from #8.

6) Take north to be +y and east to be +x.

Given:

vₓ = 2.2 m/s

vᵧ =  1.3 m/s

Find: v

v² = vₓ² + vᵧ²

v² = (2.2 m/s)² + (1.3 m/s)²

v ≈ 2.6 m/s

θ = atan(vᵧ / vₓ)

θ = atan(1.3 / 2.2)

θ ≈ 31°

7) Given:

Δy = -4.3 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)

v ≈ 9.2 m/s

8) Given:

Δy = -6.7 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.2 s

8 0
3 years ago
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