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tekilochka [14]
3 years ago
13

Help needed,

Physics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

224 J

Explanation:

Step one:

given data

Force applied = 22.4 N

distance = 10 meters

<u>Required:</u>

the work done

Step two:

The expression for the work done is

WD= force*distance

substitute

WD= 22.4*10

WD= 224 J

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Help ASAP please &amp; thank you ​
jasenka [17]

Answer:

I think its A plz tell me if im right

7 0
3 years ago
What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use
marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
3 years ago
Which of the following is most likely to happen when energy is transferred to
ankoles [38]

Explanation:

the object will begin to move

4 0
3 years ago
Read 2 more answers
Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from
svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2   force of Sun on Mars

F(Merc) = G m M / R^2    force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0
3 years ago
What does the Nucleolus do?
SCORPION-xisa [38]

Answer:

C

Explanation:

6 0
3 years ago
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