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Irina-Kira [14]
3 years ago
7

Two farmers are trying to move a large rock that sits in a field. They use a long stick as a lever and a brick as the fulcrum, a

s shown in the diagram. They apply a force on the stick at the location shown by the arrow. However, the rock does not move.
Which change would BEST allow the farmers to move the rock?
A) Moving the brick closer to the rock.
B) Moving the brick to the other side of the rock.
C) Applying a smaller force at the same location on the stick.
D) Applying an equal force at a location farther away from the rock.

Physics
2 answers:
kumpel [21]3 years ago
4 0

D is the answer for usatestprep

diamong [38]3 years ago
4 0

Answer:

D) Applying an equal force at a location farther away from the rock.

Explanation:

Here big stone is moved by applying the technique of torque

Here the applied force on the rod will give torque due to which we will have

\tau = \vec r \times \vec F

so it is product of force and its distance from fulcrum

So if the the product will be more then we will have more torque in output and that will easily shift the heavy rock

So here the torque can be increased by increasing the product of force and its distance from fulcrum

So here correct answer will be

D) Applying an equal force at a location farther away from the rock.

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Thomas is climbing Mt. Everest. What happens as he climbs farther up the mountain?
lakkis [162]

Answer:

B. the air pressure decreases

Explanation:

As elevation increases, there is less overlying atmospheric pressure mass, so that atmospheric pressure decreases with increasing elevaton.

4 0
3 years ago
Read 2 more answers
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,
valina [46]

Answer:

=2.99\times10^9J

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement S = 0.72\times10^3m

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19°   ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)

=2.99\times10^9J

6 0
3 years ago
A toroid with a square cross section 3.0 cm ✕ 3.0 cm has an inner radius of 25.1 cm. It is wound with 600 turns of wire, and it
coldgirl [10]

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T

5 0
3 years ago
The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes
Likurg_2 [28]

Answer:

The answer to the question is;

The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m  

Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

7 0
3 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi
USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,                                

length of skid = 93 m          

coefficient of friction = 0.35

final velocity = 0 m/s              

initial velocity = ?                        

force here is friction  f = μ mg

F = ma                                                

now com paring                      

-μ mg = m a                      

a = - μ g                    

a = - 0.35 x 9.8              

a = -3.43 m/s²

we know,              

v² = u² + 2 a s                        

0 = u² - 2 x 3.43 x 93                

u² = 637.98                    

u = 25.26 m/s                      

u = 25 m/s (two significant figure)

6 0
3 years ago
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