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RSB [31]
3 years ago
11

A ball is thrown at an angle of 40 degrees above the horizontal. The horizontal component of the baseball's initial velocity is

12.0 meters per second. What is the magnitude of the ball's initial velocity?
Physics
2 answers:
motikmotik3 years ago
8 0
<span> Let,
initial velocity = v m/sec,

Angle, x = 40 degrees,

horizontal-componant = v.cos(x) = 12 m/sec,
OR,
v = 12 / cos(40) = 12/0.766 = 15.67 meters/sec >================< ANSWER </span> Source(s): Fazaldin A <span> · 4 years ago </span>
Andru [333]3 years ago
7 0

Answer:

Magnitude of initial velocity = Vi = 15.7 m/s

Explanation:

Horizontal component of velocity = Vi × Cosθ = 12   ……….. (i)

Where,  

θ  = Angle at which ball is thrown above the horizontal = 40 degrees  

Put θ = 40 degrees in equation (i),

Vi × Cos(40)= 12

Vi = 12/ Cos(40)

Vi = 12/0.766

Vi = 15.66 m/s

Vi = 15.7 m/s

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Nesterboy [21]

given that snow is projected at an angle of 40 degree

It range is given as a = 19 ft

a = 19 * 0.3048 = 5.8 m

now we can use the formula of horizontal range

R = \frac{v_o^2 sin2\theta}{g}

5.8 =\frac{ v_o^2 sin(2*40)}{9.8}

5.8 = \frac{v_o^2 * sin80}{9.8}

v_o^2 = 57.7

v_o = 7.6 m/s

<u>so its initial speed must be 7.6 m/s</u>

8 0
3 years ago
A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of the car?
rodikova [14]

Answer:

16 m/s.

Explanation:

The following data were obtained from the question:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Mass of car = 2500 kg

Velocity of car =..?

Next, we shall determine the momentum of the truck. This can be obtained as follow:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Momentum of truck =.?

Momentum = mass × velocity

Momentum = 5000 × 8

Momentum of the truck = 40000 Kg.m/s

Finally, we shall determine the velocity of the car as follow:

From the question given above, we were told that the car and truck has the same momentum.

This implies that:

Momentum of the truck = momentum of car = 40000 Kg.m/s

Thus, the velocity of the car can be obtained as shown below:

Mass of car = 2500 kg

Momentum of the car = 40000 Kg.m/s

Velocity of car =..?

Momentum = mass × velocity

40000 = 2500 × velocity

Divide both side by 2500

Velocity = 40000/2500

Velocity = 16 m/s

Therefore, the velocity of the car is 16 m/s.

5 0
3 years ago
Suppose two wagons, one with twice as much mass as the other, fly apart when a compressed spring that joins them is released. Th
sp2606 [1]

Answer:

The heavier wagon rolls 1/2 as fast as the lighter wagon.

Explanation:

When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration  will become half of the heavier wagon in comparison with lighter one.

7 0
3 years ago
An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 35
LekaFEV [45]

Answer:

Weight = 966 Newton.

Explanation:

Given the following data;

Length = 1.2 m

Width = 2.3 m

Pressure = 350 Pa

To find the weight of the tank;

We know that weight is the force of gravity acting on an object multiplied by its mass.

Weight = mg = force

Hence, we would determine the force using the parameters that were given.

But we would first determine the area of the rectangular tank.

Area of rectangle, A = length * width

A = 1.2 * 2.3

A = 2.76 m²

Mathematically, pressure is given by the formula;

Pressure = force/area

Force = pressure * area

Substituting into the formula, we have;

Force = 2.76 * 350

Force = 966 Newton

Therefore, the weight of the tank is 966 Newton.

5 0
3 years ago
A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300
Illusion [34]
<h2>Answer:</h2>

\boxed{P=96.09MW}

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega

So we can calculate the power loss as:

P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}

Finally, the power loss due to resistance in the line is 96.09MW

5 0
3 years ago
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