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dem82 [27]
3 years ago
14

How can you increase the torque you are putting on a lever, without exerting any more force?

Physics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

See below

Explanation:

Torgue = f * distance    basically ...so if you want to increase torque with same  f   then the distance ( lever arm)  must be increased

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An object of mass 1.5 kg rests on a shelf where it has a gravitational potential energy of 7 joules. An object of mass 4.5 kg is
bija089 [108]

Answer:

C. 21 Joules

Explanation:

We apply the formula to calculate the potential energy (Ep):

Ep=m*g*h

Where:

Ep : potential energy in Joules (J)

m :mass in kilograms (kg)

g  acceleration due to gravity  (m/s²)

h: height in meters (m)

Calculation of the height (h)

Ep = m*g*h

7 = (1.5 )*(9.8) *(h )

7 = (14.7) (h )

h = 7 /  (14.7)

h= 0.476 m

Gravitational potential energy of the second object

Ep = m*g*h

Ep = (4.5 )*(9.8) *(0.476 )

Ep = (4.5 )*(9.8) *(0.476 )

Ep = 21 J

6 0
4 years ago
Block 1, of mass m1 = 1.30 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mas
Dennis_Churaev [7]

(A) The total initial momentum of the system is

(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s

(B) Momentum is conserved, so that the total momentum of the system after the collision is

35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>

where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives

<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s

(C) The kinetic energy of the system after the collision is

1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J

and before the collision, it is

1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J

so that the change in kinetic energy is

∆<em>K</em> = 25.4 J - 474 J ≈ -449 J

6 0
3 years ago
HELP!!!!
kykrilka [37]

Answer:

Clouds form when below the dew point

4 0
3 years ago
A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to
o-na [289]

Answer:

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

to find out

angle between the wrench handle and the direction of the applied force

solution

we will apply here torque equation that is express as

torque =  distance × force × sin(θ)   ...................1

put here value we will get angle that is

15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

6 0
3 years ago
Read 2 more answers
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet
Shtirlitz [24]

The average distance between the planet and the star is:

R=9.36*10^11 m

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

As was already said, Earth is located roughly 150 million kilometres (93 million miles) from the Sun on average. It is 1 AU. Mars is on our fictitious football field's three-yard line. On average, the distance between the Sun and the red planet is around 142 million miles (228 million kilometres).

Learn more about average distance:

brainly.com/question/18366547

#SPJ4

The complete question is ''If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 10^30 kg, what is the average distance between the planet and the star? 1.43 × 10^9 m 9.36 × 10^11 m 5.42 × 10^13 m 9.06 × 10^17 m''.

4 0
1 year ago
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