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3 years ago

9

A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity

of the ball

1 answer:

raketka [301]3 years ago

8 0

25 m/s is the answer

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A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r

andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature $T_1=781 K$

Reservoir 2 Temperature $T_2=335 K$

Let Q is the amount of heat Flows i.e. $Q=1477 J$

thus change in Entropy is given by $\frac{\sum Q}{T}$

$\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}$

$\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}$

$\Delta S=\frac{\sum Q}{T}=-1.891+4.4089$

$\Delta S=\frac{\sum Q}{T}=2.517 J/K$

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3 years ago

A river has a steady speed of 0.550 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. If

nlexa [21]

Answer:

$t=564.83seconds=9.414minutes=0.1569hours$

Explanation:

Here the steady speed of the river water relative to ground is

$V_{wg}=0.55m/s$

Speed of the boy relative to still water is

$V_{BW}=1.50m/s$

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

$V_{BGup}=V_{BW}-V_{WG}\\V_{BGup}=1.50m/s-0.550m/s\\V_{BGup}=0.95m/s$

And the speed of boy relative to the ground downstream is:

$V_{BGdown}=V_{BW}+V_{WG}\\V_{BGdown}=1.50m/s+0.550m/s\\V_{BGdown}=2.05m/s$

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

$t=d/v$

The total time taken for one trip is:

$t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours\\$

3 0

3 years ago

A submarine that is in the sea what principle is that

SIZIF [17.4K]

A U-boat was a german submarine

4 0

3 years ago

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert

zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

$F=\dfrac{mv^2}{r}$

v is the maximum speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s$

Hence, the maximum speed of the ball is 5.86 m/s.

3 0

3 years ago

The ________ of a gas can affect the pressure of a gas.

Diano4ka-milaya [45]

d. none of the above

pressure affects volume not the other way round

5 0

3 years ago

Read 2 more answers