1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andru [333]
3 years ago
7

Give an example in which a small force exerts a large torque. give another example in which a large force exerts a small torque.

Physics
2 answers:
algol [13]3 years ago
5 0

Answer:

Explanation:

Torque is defined as the product of force and the perpendicular distance.

If we take a screw driver of small length then we need to apply a large force to open or tighten the screw.

If a screw driver of long arm, then we need to apply small force to open or tighten the screw.

goldenfox [79]3 years ago
3 0
<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.

With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
</span><span>
</span>
You might be interested in
A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.
Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m

The maximum height above the ground that the ball reaches is 7.09683 m

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s

The velocity just before it hits the ground is 11.8 m/s

6 0
3 years ago
5. What happens to the arrangement of water molecules as ice melts?
m_a_m_a [10]

Answer: I am pretty sure the answer is B

Explanation: If not sorry bro.

7 0
3 years ago
PLEASE HELP ANSWER ASAP. 10 points. I will give brainliest.
alina1380 [7]

The answer is D. I know because I already answered the question.

3 0
3 years ago
A mass of 5kg accelerates at 3m/s/s, how much force was put on it?
goldfiish [28.3K]

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

5 0
3 years ago
Read 2 more answers
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
3 years ago
Other questions:
  • Smallest unit that consists of at least one atom
    7·2 answers
  • Which of the following is not affect by muscles a blood temperature b digestion c hair growth d speech
    6·1 answer
  • What is E=m² the famous theory of albert einstien?
    12·2 answers
  • A solution is oversaturated with solute. which could be done to decrease the oversaturation?
    11·2 answers
  • In a lab, four balls have the same velocities but different masses.
    6·2 answers
  • How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
    11·1 answer
  • Of the three types of radiation listed below, which is the least penetrating and which is the most penetrating? -alpha -beta -ga
    8·1 answer
  • How many degrees equal π in radians? How many revolutions (turns) equal π radians? ​
    10·1 answer
  • A 1kg mass is moving at 1m/s. What is its kinetic energy?
    12·1 answer
  • A 2kg mass is moving at 3m/s. What is its kinetic energy?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!