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3 years ago

If you cannot exert enough force to loosen a bolt with a wrench, which of the following should you do?

Physics

1 answer:

stepan [7]3 years ago

6 0

Answer:

2.) Use a wrench with a longer handle.

Explanation:

It is suggested to use a wrench with longer handle if you cannot exert enough force to loosen a bolt with a wrench.

In doing this work, the torque which is the force that can turn the screw is not enough;

Torque = force x distance

If we increase the distance or the length of the handle, we can generate more torque to overcome the force needed.

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Marcus attends a night program at an observatory to learn more about the solar system. He learns that the solar system contains

Lesechka [4]

Answer: D meteoroid

Explanation:

Meteoroid is the smallest among them. It is a tiny asteroid or the broken-off crumb of comets and sometimes planets. It ranges in size from a grain of sand to boulders 3 feet (1 meter) wide. When meteoroids collide with a planet's atmosphere, they become meteors. If those meteors survive the atmosphere and hit the planet's surface, their remains are called meteorites

8 0

3 years ago

What is the distance of a car travels is its velocity is 30 mph and its time is 2 hours ?

Alborosie

Answer:

distance = 60 miles

Explanation:

3 0

2 years ago

Atmospheric pressure varies from day to day. The level of a floating ship on a high-pressure day is (a) higher (b) lower, or (c)

frez [133]

Answer:

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.

<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>

5 0

3 years ago

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,

Aloiza [94]

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = $\frac{mv^{2} }{r}$

where velocity (v) = $\frac{2πr}{t}$

tension now becomes (T) = $\frac{m}{r} x (\frac{2πr}{t})^{2}$

tension (T) = $\frac{4π^{2}rm }{t^{2} }$

now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = $\frac{4π^{2}x0.5x0.25}{0.5^{2} }$

tension (T) = $2π^{2}$ = $2x3.142^{2}$ = 19.74 N

4 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$