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3 years ago

Nancy is pushing her empty grocery cart at a rate of 1.8 m/s. 30 seconds later,

Physics

1 answer:

Strike441 [17]3 years ago

3 0

3.Es tarde y mi taxi no llega. Estoy ____.
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(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from

victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c) $a_{satellite}= 3.13 m/sec^2$

d) $a_{earth}=2.32\times10^{-22} m/s^2$

Explanation:

The force experienced by the satelite is giveb by

$F= \frac{Gm_{satellite}m_{earth}}{r^2} \\$

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

$F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}$

⇒F= 1396.52 N

now,

$a_{satellite}= \frac{F}{m_{satellite}}$

$a_{satellite}= \frac{1396.52}{445}$

$a_{satellite}= 3.13 m/sec^2$

also,

$a_{earth}= \frac{F}{m_{earth}}$

$a_{earth}= \frac{1396.52}{(6\times10^24)}$

$a_{earth}=2.32\times10^{-22} m/s^2$

3 0

3 years ago

If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit

Oxana [17]

Answer:

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

4 0

3 years ago

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago

What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?

ra1l [238]

A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour

Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

8 0

3 years ago

S waves arrive at distant points before other seismic waves, true or false

Step2247 [10]

False. p waves, than s waves

3 0

3 years ago