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3 years ago

I WILLL MARK YOUUUUUUU HELPPPP

Physics

2 answers:

Ksivusya [100]3 years ago

3 0

Answer:

10.75 kg

Explanation:

moon = (weight on earth/9.81m/s^2)*1.622m/s^2

Firlakuza [10]3 years ago

3 0

Answer:

weight of the moon = mass × gravity

= 65kg × 1•6

=104N/kg

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A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?

Tatiana [17]

Answer:

<u>We are given:</u>

displacement (s) = 130 m

acceleration (a) = -5 m/s²

final velocity (v) = 0 m/s [the cars 'stops' in 130 m]

initial velocity (u) = u m/s

<u>Solving for initial velocity:</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (u)² = 2(-5)(130)

-u² = -1300

u² = 1300

u = √1300

u = 36 m/s

8 0

3 years ago

What is precision?what is precision?

RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

3 years ago

A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

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3 years ago

Can I PLEASE get some help? I REALLY need it!

soldi70 [24.7K]

The answer is C. Hope this helps.

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3 years ago