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3 years ago

I NEED THIS ANSWER TODAY PLEASE HELP ME

2 answers:

7 0

The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.

Inessa05 [86]3 years ago

5 0

B. The time of day

^^ not really sure but haha

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With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why?

mina [271]

Answer:C: The gravity on the moon is less than the gravity on Earth.

Explanation:

4 0

3 years ago

Suppose that you are standing on a train accelerating at 0.39g. What minimum coefficient of static friction must exist between y

timurjin [86]

Answer:

0.39

Explanation:

In order not to slide, you must have exactly the same acceleration of the train:

$a=0.39 g$

where

g = 9.81 m/s^2

There is only one force acting on you: the static frictional force that "pulls" you forward, and it is given by

$F_s = \mu_s mg$

According to Newton's second law, the net force acting on you (so, the frictional force) must be equal to your mass times the acceleration, so we have

$F= ma = \mu_s mg$

from which we find

$\mu_s = \frac{a}{g}=\frac{0.39 g}{g}=0.39$

so, the minimum coefficient of static friction must be 0.39.

5 0

3 years ago

How is high temperature achieved by concave mirror?

tensa zangetsu [6.8K]

A concave mirror is used in the design of solar furnaces because they converge the parallel sunrays at a point. This helps to increase the temperature of the furnace.

3 0

2 years ago

Read 2 more answers

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between the

Sever21 [200]

Answer:r=5.824 mm

Explanation:

Given

Charge $q_1=70 nC$

$q_2=70 nC$

Force between them $F=1.30 N$

Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$

5 0

3 years ago

A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17

Korolek [52]

Given:

Mass of cart: 15kg

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

So, the solution would be like this for this specific problem:

1) W(by applied force) = F(applied) x s x cosθ 
=>W(a) = 12 x 14 x cos17* = 160.66 J 

2) By F(net) = F(applied) - F(friction) 
=>As v = constant => a = 0 => F(net) = 0
=>F(applied) = F(friction)
=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s
=>W(f) = -12 x 14 x cos17* = -160.56 J 

3) 0, as the displacement is perpendicular to Force 

4) 0, as the displacement is perpendicular to Force

To add, the force that is applied to an object by a person or another object is called the applied force.

8 0

3 years ago