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Bogdan [553]
3 years ago
9

25 points!

Physics
1 answer:
pentagon [3]3 years ago
3 0

Answer:

6N

Explanation:

Given parameters:

Mass of object  = 6kg

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time  = 30s

Unknown:

Net force acting on the object  = ?

Solution:

From Newton's second law of motion:

   Force  = mass x acceleration

Acceleration is the rate of change of velocity with time

  Acceleration  = \frac{Final velocity  - Initial velocity }{time}  

  Force  = mass x  \frac{Final velocity  - Initial velocity }{time}  

So;

 Force  = 6 x \frac{25  - 5}{30}    = 6N

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Charges can be separated in an isolated system
Vesnalui [34]

Answer:

yes. can always be separated

4 0
3 years ago
A sound system is being set up in a gazebo in a park. It needs to produce music so that everyone can hear it. How much power wou
Agata [3.3K]

Answer:

Power of the speakers = 1.57 \times 10^{-6} W

Explanation:

There are speakers setup in a park and we have to find the power it would require to have intensity of sound to be 10^{-8} W/m^2 at a distance of 5 meter.

In one dimension the intensity is constant as the wave travels. In two or three dimensions, however, the intensity decreases as you get further from the source.

Intensity is inversely proportional to the square of the distance from the point of sound production.

Here area = 2\times \pi \times r^{2}

area = 157 m^{2}

Intensity = \frac{Power}{Area}

Power = intensity \times area

Power = 10^{-8} \times 157Power = 1.57 [tex]\times 10^{-6} W

7 0
3 years ago
A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj
Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area  

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = \sigma_B= \frac{M_y}{I} where I = \frac{(0.5^2 + 4^2)}{12}

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0
3 years ago
Read 2 more answers
Collision practice
NikAS [45]

Answer:

v_f=4\:\mathrm{m/s\: in\:Biff's\:direction}

Explanation:

Using the Law of Conservation of Momentum, we can set up the following equation:

m_2v_2-m_1v_1=m_fv_f, where m_1v_1 is Bruce's momentum and m_2v_2.

Plugging in given values, we get:

90\cdot7-45\cdot2=(90+45)v_f, \\v_f=\frac{540}{135}=\fbox{$4\:\mathrm{m/s}$}.

6 0
3 years ago
A particle is located on the x axis 4.9 m from the origin. A force of 38 N, directed 30° above the x axis in the x-y plane, acts
Juli2301 [7.4K]

Answer:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

        Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

7 0
3 years ago
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