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IgorLugansk [536]
3 years ago
5

According to Hooke's law

Physics
1 answer:
SOVA2 [1]3 years ago
7 0

Answer:

According to Hook's law, we know,

strain/stress =Constant

Explanation: So, the ratio between stress and strain is always constant.

So, if stress is increased, then strain changes in that way so that this ratio always remains constant.

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Compare the gravity between these pairs, each consisting of an Earth-like planet and its star. You are given the mass of the pla
Maru [420]

Answer:

The answer is below

Explanation:

Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:

F=G\frac{m_1m_2}{r^2} \\\\Where\ F=force,G=gravitational\ constant, m_1\ and\ m_1=mass\ of\ objects,r\ =distance\ between \ the\ two\ objects.

The masses and distances for this question is in common units, Therefore the result would be in ratios

a) 4 MEarth / 2 MSolar / 3 AU

The force (F) = (4 * 3) / 3² = 4/3

b) 1 MEarth / 1 MSolar / 1 AU

The force (F) = (1 * 1) / 1² = 1

c) 1 MEarth / 2 MSolar / 2 AU

The force (F) = (1 * 2) / 2² = 1/2

6 0
3 years ago
while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o
sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}

The minimum radius not to exceed the centripetal acceleration is 2401 m.

8 0
2 years ago
Austin buys a new moped he travel 3 km south and then 4 km east and then 3km north how far does he need to go to get back to whe
Ira Lisetskai [31]
The first thing to do is to define the origin of the coordinate system as the point at which the moped journey begins.
 Then, you must write the position vector:
 r = -3j + 4i + 3j
 Rewriting
 r = 4i
 To go back to where you started, you must go
 d = -4i
 That is to say, must travel a distance of 4Km to the west.
 Answer
 West, 4km.
4 0
3 years ago
The field-line representation of the e-field in a certain region in space is shown below. The dashed lines represent equipotenti
lorasvet [3.4K]

You need to go on google an dlook this up

3 0
3 years ago
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
3 years ago
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