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3 years ago

If a ball is launched horizontally at 40 m/s

Physics

1 answer:

vaieri [72.5K]3 years ago

4 0

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

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The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com

Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

$Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\$

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

$Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\$

<u>Direction = 65.48°</u>

4 0

3 years ago

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to t

snow_lady [41]

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

$m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}$ ..........(1)

$m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s$

Substituting the above values into equation (1) and make V2 the subject of the formula:

$0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\$

$V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}$ ..................(2)

Using the law of conservation of kinetic energy:

$0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\$

$2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}$ .......(3)

Substitute equation (2) into equation (3)

$2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s$