A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?

Which of the following statements correctly describes the models put forth by Ptolemy and Copernicus to explain the universe? A.

FrozenT [24]

The correct option is D.
The model developed by Ptolemy has a lot of inconsistency and during the middle age additional explanation was offered for the claims made by the model. The model was very complicated because it was based on erroneous assumptions.
Copernicus model was simpler and some of his claims were correct.<span />

8 0

2 years ago

Read 2 more answers

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

8 0

1 year ago

Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex

Tpy6a [65]

Can you attach a picture of the actual problem?

7 0

2 years ago

Read 2 more answers

8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

hence, B_o = 1.013μT

6 0

2 years ago