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2 years ago

Which scientist discovered the laws that determine the motion of a spring?

Physics

1 answer:

Kruka [31]2 years ago

7 0

Answer: Robert Hooke

Explanation: Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.

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What is the weight of a dog with a mass of 12 kg? Explain how you obtained your answer.

butalik [34]

Answer: 117.60N

Explanation:

Weight is a force. Therefore, we can use the force formula to find weight.

$Formula: W=m*g$

W = weight

m = mass

g = acceleration due to gravity ( $9.8m/s^2$ )

$W=(12kg)(9.8m/s^2)\\W=117.60N$

5 0

4 years ago

Read 2 more answers

What is the magnitude of the gravitational force acting on the earth due to the sun?

expeople1 [14]

Answer: $3.524(10)^{22}N$

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m1=1.99(10)^{30}kg$ is the mass of the Sun

$m2=5.972(10)^{24}kg$ is the mass of the Earth

$r=1.50(10)^{11}m$ is the distance between the Sun and the Earth

Substituting the values in (1):

$F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}$ (2)

Finally:

$F=3.524(10)^{22}N$ This is the gravitational force acting on the earth due to the sun

3 0

3 years ago

Read 2 more answers

A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af

Lady_Fox [76]

Answer:

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

4 0

3 years ago

A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What en

Gelneren [198K]

Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) $U=668.16\ J$

Explanation:

Given:

mass of the child, $m=24\ kg$

height of the slide, $h=3.3\ m$

initial velocity of the child at the slide, $v_i=0 m.s^{-1}$

final velocity of the child at the bottom of slide, $v_f=3\ m.s^{-1}$

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

$PE=m.g.h$