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oksano4ka [1.4K]
3 years ago
6

If I have a 10 lb ball and a 2 lb ball which

Physics
1 answer:
mojhsa [17]3 years ago
6 0

Answer:

It would be the 10 lb ball.

Explanation:

The more that an object weighs, the more force that is needed to accelerate it.

Hope this helped

:)

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Help please (20 pts) I think it's C
gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

6 0
3 years ago
A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car​
Vsevolod [243]

Answer:

960 m

Explanation:

Given that,

  • Speed = 120 m/s
  • Time taken = 4 minutes

We have to find the distance covered.

Firstly, let's convert time in seconds.

→ 1 minute = 60 seconds

→ 4 minutes = (4 × 60) seconds

→ 4 minutes = 240 seconds

Now, we know that,

→ Distance = Speed × Time

→ Distance = (4 × 240) m

→ Distance = 960 m

Therefore, distance covered is 960 m.

7 0
3 years ago
ILL GIVE BRAINLYEST
lawyer [7]

Answer:

3rd picture straight line going up right

Explanation:

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5 0
4 years ago
a child in a tree house uses a rope attached to a basket to lift a 26 nn dog upward through a distance of 5.0 mm into the house.
agasfer [191]

Total work done is 0.13 Joules

<h3>What is work done ?</h3>

The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.

According to the given information

We need to find the work done

work done  = force × distance

We are given,

force  = 26 N

Distance = 0.0005 meter

hence ,

Work done  = 26 × 0.005  

                   = 0.13 Joules

Total work done is 0.13 Joules

To know more about Work done

brainly.com/question/13662169

#SPJ4

4 0
1 year ago
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi
True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0
3 years ago
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