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Setler [38]
3 years ago
13

How deep can an object with 6360N hitting on a sponge get ???

Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

The sponge must go \dfrac{636}{m}\ \text{meter} deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}

So, the sponge must go \dfrac{636}{m}\ \text{meter} deep.

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Why might a balloon, that is inflated almost to its capacity, pop or explode on an extremely warm day?
REY [17]
On an extremely warm day, the balloon might pop because gases expand the hotter they get, and due to its temperature it is likely to pop if it is, indeed, nearly, if not completely, filled to its capacity.  I hope this helps, have a nice day!
7 0
3 years ago
Read 2 more answers
Can you please tell me what this is
Anna [14]

Answer:

200000 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE => is the kinetic energy.

m =>is the mass of the object

V => it the velocity of the object.

With the above formula, we can obtain the kinetic energy of the roller coaster as follow:

Mass (m) of roller coaster = 1000 Kg

Velocity (v) of roller coaster = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 1000 × 20²

KE = 500 × 400

KE = 200000 J

Therefore, the kinetic energy of the roller coaster is 200000 J.

4 0
2 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
What force must be used to do 224 Joules of work on an object over a distance of 32 meters?
cestrela7 [59]
7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons
4 0
3 years ago
Using 1000 j of work, a toy elevator is raised from the ground floor to the second floor in 20 s. how much power does the elevat
Anastasy [175]
Power (watts) = work done (joules) ÷ time (seconds)

power (watts) = 1000J ÷ 20s

power (watts) = 50w

3 0
2 years ago
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