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sleet_krkn [62]
3 years ago
13

A high diver dives off a platform and hits the water 3.6 seconds later. How high was the platform?

Physics
2 answers:
xenn [34]3 years ago
6 0

Answer:

A high diver dives off a platform and hits the water 3.6 seconds later. How high was the platformExplanation:

HACTEHA [7]3 years ago
3 0

Answer:

33m

Explanation:

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What force is the total force felt by an object?
skelet666 [1.2K]

Answer:

net force

Explanation:

Net force felt by an object.

6 0
3 years ago
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Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t
umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2>F = 1991.05 N</h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

mg \times L = F \times d

so we have

18.14 \times 9.8 \times 0.56 = F \times (0.05)

F = 1991.05 N

8 0
3 years ago
What is the potential energy of a 5kg rock that is 7m high on a hill
Goryan [66]

Explanation:

Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.

4 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be
ale4655 [162]

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

  • The energy at rest which is the potential energy is 500J
  • This energy will be converted to kinetic energy in total after the arrow has been released.
  • This way, no energy is lost and we can account for the energy transformations occurring.
4 0
3 years ago
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