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3 years ago

The gradual temperature change from one season to the next is due to what? view attachment

Physics

1 answer:

Olenka [21]3 years ago

3 0

Global warming jk part of the earth is either facing the sun or part of the sun or away from the sun

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Mercury is in the 80th position in the periodic table. How many protons does it have?

Verdich [7]

Mercury has 80 protons. Ironic?

7 0

3 years ago

Read 2 more answers

what is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes

adell [148]

Answer:

explain this

Explanation:

7 0

3 years ago

A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien

Liula [17]

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

=ma

Solve for a=4.19m/s^2

7 0

3 years ago

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&

Katen [24]

Answer:

a) T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 = $\frac{T_x}{T}$

cos 30 = $\frac{T_y}{T}$

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m $\frac{v^2}{r}$ (2)

a) we substitute in 1

T = $\frac{mg }{cos 30}$

T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

T = 2.26 N

b) from equation 2

v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

v = 1.68 m / s

8 0

3 years ago

QUESTION 7

ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

$v_f=v_o+at$

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

$\displaystyle a=\frac{0-20.4}{7.4}$

$a=-2.757\ m/s^2$

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

$F = 1,126\ kg * 2.757\ m/s^2$

F= 3,104 N

The force required is 3,104 N

6 0

3 years ago