Please I need HELP WITH THIS QUESTION!!!

Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he

storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0

3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface

Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

$g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}$

The expression of escape velocity is given by

$v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s$

The escape speed is 1777.92 m/s

3 0

3 years ago

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QUESTION 7

ryzh [129]

Answer:

The force required is 3,104 N

Explanation:

Force

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

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

How is the moment magnitude scale used to describe earthquakes?

natka813 [3]

The Richter Scale is not commonly used anymore, except for small earthquakesrecorded locally, for which ML and Mblg are the only magnitudes that can be measured. For all other earthquakes, the moment magnitude scale is a more accurate measure of the earthquake size.

3 0

3 years ago

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