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allsm [11]
2 years ago
12

A 50 N object is on Earth. What is its mass?

Physics
1 answer:
faust18 [17]2 years ago
7 0

Answer:   5.10 KG

Explanation:  W = 50 N

                       BY FORMULA

                         W = MG

                          M = W/G

                          M = 50 /9.8

                          M=  5.10 KG

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Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and th
Firdavs [7]

Answer:

Torque = 35.60 N.m (rounded off to 3 significant figures.

Explanation:

Given details:

The mass of the rock on the left, ms = 2.25 kg

The total mass of the rocks, mp = 10.1 kg

The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m

(a) The torque produced by the pile of rock, T = F*rp = m*g*rp

Torque = 9.8*0.360*10.1 = 35.6328

Torque = 35.60 N.m (rounded off to 3 significant figures).

5 0
3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
2 years ago
Read 2 more answers
A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati
Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is 5 N

Tray makes an angle of \theta =12^{\circ}

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force W_p=W\cos \theta

W_p=5\times \cos 12=4.89 N

Along the Tray

W_{along}=W\sin \theta

W_{along}=5\times \sin 12=1.04 N

Thus 1.04 N is the magnitude of force that will cause Platter to slide down  

7 0
3 years ago
A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff
jekas [21]

Answer:

F = 505.13 N

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

F_n + Fsin60 = mg

here we know

m = 180 lb

m = 81.65 kg

F_n = 81.65(9.81) - Fsin60

F_n = 801 - 0.866 F

now friction force on the crate is given as

F_x = \mu F_n

Fcos60 = 0.7(801 - 0.866 F)

0.5F + 0.61F = 560.7

F = 505.13 N

Yes it is better to pull the rope rather than push it

6 0
3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

8 0
3 years ago
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