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3 years ago

15

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

f the hockey puck is true?
The hockey puck will come to a halt immediately.

The hockey puck will continue moving with constant velocity along a straight line.

The hockey puck will accelerate and continue moving in straight line.

The hockey puck will continue moving in random direction.

2 answers:

77julia77 [94]3 years ago

8 0

C. the hockey puck will accelerate and continue moving in a straight line

Llana [10]3 years ago

7 0

Answer:

The hockey puck will continue moving with constant velocity along a straight line.

Explanation:

As per Newton's first law we know that when there is no net force on an object then the velocity of object will always remains constant

Or we can say that the acceleration of an object will be zero when the force applied on the object is zero.

As per Newton's law we know

$F = ma$

so we will have

a = 0

so object will continue in straight line with same uniform speed without any change in magnitude as well as any change in direction

so correct answer will be

The hockey puck will continue moving with constant velocity along a straight line.

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A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

Lady_Fox [76]

Answer:

The magnetic field strength needed is 1.619 T

Explanation:

Given;

Number of turns, N = 485-turn

Radius of coil, r = 0.130 m

time of revolution, t = 4.17 ms = 0.00417 s

average induced emf, V = 10,000 V.

Average induced emf is given as;

V = -ΔФ/Δt

where;

ΔФ is change in flux

Δt is change in time

ΔФ $= -NBA(Cos \theta_f - Cos \theta_i)$

where;

N is the number of turns

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle of inclination of the coil and the magnetic field,

$\theta_f = 90^o\\\theta_i = 0^o$

V = NBACos0/t

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

5 0

3 years ago

Read 2 more answers

If you are in a car at a stop sign and then the driver

SVEN [57.7K]

Answer:

it wont be because the hand brake would be on

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3 years ago

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What are three mechanisms for reproductive isolation? Which Mechanism isolates two populations of similar frogs with different m

saul85 [17]

The mechanisms of reproductive isolation are:
1) Habitat isolation
2) Behavioral isolation
3) Mechanical isolation
4) Gametic isolation
The isolation happening here is behavioral isolation, because their different behavior patterns prevent the frogs from breeding.

4 0

3 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

$F_c=\frac{mv^{2} }{R}$

If we substitute the expression for $v$ in this formula, we get:

$F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}$

Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ( $G=6.67*10^{-11} Nm^{2}/kg^{2}$ ) and M is the mass of the Earth ( $M=5.97*10^{24}kg$ ). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

$R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km$

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0

3 years ago

A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon

jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

Στ = 0

W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

x₂ = $\frac{M_1g }{m_2 g} \ x_1$

let's calculate

x₂ = $\frac{100}{75} \ 1$

x₂ = 1.33 m

7 0

3 years ago