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3 years ago

11

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo

nent of the ball? Round the answer to the nearest tenth of a m/s

2 answers:

Sever21 [200]3 years ago

8 0

The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle.
x-component of speed = (31 m/s)(cos 35°)
= 25.39 m/s
Thus, the horizontal velocity component of the ball is 25.39 m/s.

lawyer [7]3 years ago

5 0

Answer:

Horizontal component of the speed is given as

$v_x = 25.4 m/s$

Explanation:

As we know that if a vector is inclined at some angle with the horizontal then the component of the vector along and perpendicular to horizontal is given as

$A_x = Acos\theta$

$A_y = Asin\theta$

now here we need to find the component of velocity

so we know that

$v = 31 m/s$

$\theta = 35^0$

now we will have

$v_x = 31cos35$

$v_x = 25.4 m/s$

so horizontal component of the speed will be 25.4 m/s

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The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following

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Time takes by the moon to go through all the phases

about 29.5 days

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At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!

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3 years ago

Read 2 more answers

Find the potential energy of a 2 kg ball 15 m in the air.

mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

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3 years ago

A softball player moving 3.89 m/s

Ahat [919]

Answer:

0.119 s

Explanation:

Given that

$U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m$

Lets take final speed of the softball after covering 4.8 m = V

We know that

$V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s$

Also We know that

$V=U+at\\$

Putting the value of V ,U and a in the previous equation We get

$3.7=3.89-1.44\times t\\t=0.119\ s$

Therefore slide time will be 0.119 s

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4 years ago

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

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<u>Given </u><u>:</u><u>-</u>

An elevator is moving vertically up with an acceleration a.

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

The force exerted on the floor by a passenger of mass m .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>

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3 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

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2 years ago