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2 years ago

14

Is torque only produced when the force is applied perpendicular to the moment arm?

2 answers:

trasher [3.6K]2 years ago

8 0

Yes. Only a perpendicular component produces a torque.

Zanzabum2 years ago

6 0

Answer:yes

Explanation:

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The weight of a body of certain mass becomes zero in space.why?write with reasons

Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0

2 years ago

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

$mg(H-h)=\frac{1}{2}mv^2$

$g(H-h)=\frac{1}{2}v^2$

$\Rightarrow H=\frac{v^2}{2g}+h$

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

$=v\frac{2g}{h}$

$\Rightarrow v^2=\frac{gD^2}{2h}$

thus, height H in terms of D is given by

$H=\frac{D^2}{2h}+h$

$H=\frac{13.9^2}{2\times3}+3$

H=35.20 m

4 0

3 years ago

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How

allsm [11]

Answer:

D

Explanation:

6/35=0.17

8 0

3 years ago

What force is needed to move a 5 kg mass with an acceleration of 5 m/s²?

ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

4 0

1 year ago

Read 2 more answers

The half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su

neonofarm [45]

Answer:

(a) $N=19\times e^{-\lambda t}$

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

$\lambda =\frac {0.6931}{T}$

The equation of radioactivity used here is given by

$N=N_{o}e^{-\lambda t}$

$N=19\times e^{-\lambda t}$

(b) N = 8 gram

Substitute the values in above equation

$\lambda =\frac {0.6931}{12}$

λ = 0.0577 per hour

So, $8=19\times e^{-0.577t}$

$e^{-0.0577t}=0.421$

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

4 0

3 years ago