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During a race, four competitors of the same weight rode identical bicycles for 10 minutes. At 8 minutes, which bicycle was movin

Sophie [7]

Answer:

All the competitors will move with the same velocity.

Explanation:

Here, the situations for each competitor are identical. Thus, they will exert the same force and hence, their velocities at each instants will be identical.

6 0

3 years ago

Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti

Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

7 0

3 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

3 years ago

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

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

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

2 years ago

Two astronauts in space with a baseball decide to play catch to pass the time. In the language of conservation of momentum, desc

Anna35 [415]

As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).

4 0

3 years ago

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