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Marina CMI [18]
3 years ago
8

A marble is thrown horizontally with a speed of 7 m/s. When the marble lands, it hastraveled a horizontal displacement of 30 m.

How much time was the marble in the air?
Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

t = 4.28 s

Explanation:

The horizontal speed of a marble = 7 m/s

The horizontal displacement covered by the marble = 30 m

We need to find the time for which the marble is in air. Let the time is t. Using the formula for speed to find it as follows :

s=\dfrac{d}{t}\\\\t=\dfrac{d}{s}\\\\t=\dfrac{30\ m}{7\ m/s}\\\\=4.28\ s

So, it will in the air for 4.28 s.

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As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horiz
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solution:

long to travel 19.6m at 28m/s

t =19.6/28

=time taken to drop from rest

s=ut+(1/2)at^2

u=0

a=~10

so s=(1/2)*10*(19.6/28)^2 =2.45m

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A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
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Answer: 2.5N

Explanation:

Given the following :

Mass of block (m) = 2kg

Coefficient of static friction (μs) = 0.4

Horizontal force applied to the block = 2.5N

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First calculate the maximum static frictional force:

Frictional force = Coefficient of static friction(μs) × normal reaction(R)

Normal reaction(R) = mass × acceleration due to gravity (10m/s^2)

R = 2 × 10 = 20

Fmax = μs × R

Fmax = 0.4 x 20 = 8N

Here, since the applied force (2.5N) is less than maximum frictional force(8N).

The force of friction between the block and the floor will be equal to the applied force of 2.5N due its ability to adjust itself in other to ensure equilibrium.

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Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius,
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Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

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A 4kg box accelerated from rest by a force across the floor at a rate of 2m/s^2 for 7 seconds. What is the net work done on the
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Read 2 more answers
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider compl
Novosadov [1.4K]

Answer:

T = 2.82 seconds.

The frequency \mathbf{f = 0.36 \ Hz}

Amplitude A = 25.5 cm

The maximum speed of the glider is \mathbf{v = 56.87 \ rad/s}

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

T = \frac{31}{11}

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

f = \frac{1}{T}

f = \frac{1}{2.82}

\mathbf{f = 0.36 \ Hz}

The amplitude is determined by using the formula:

A = \frac{d}{2}

The limits that the spring makes the oscillations are from 10 cm to 61 cm.

The distance of the glider is, d = (61 - 10 )cm = 51 cm

Replacing 51 for d in the above equation

A = \frac{51}{2}

A = 25.5 cm

The maximum speed of the glider is:

v = A \omega

where ;

\omega = \frac{2 \pi}{T}

\omega = \frac{2 \pi}{2.82}

\omega = 2.23 \ rad/s

v = A \omega

v = 25.5 *2.23

\mathbf{v = 56.87 \ rad/s}

3 0
3 years ago
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