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Anna11 [10]
4 years ago
9

How much force, in g cm/s2

Physics
2 answers:
scoray [572]4 years ago
3 0
In this question force is measured in  g cm/s2 so we know that to get the answer we times g by cm/s2 
50 × 20 = 1000
Vinvika [58]4 years ago
3 0

Answer:

F=1000\frac{gr \times cm}{s^{2} }

Explanation:

Givens

m=50g

a=20cm/s^{2}

To find the force exerted by the golf ball, we need to use Netwon's second law, which is defined as

F=ma

In words, when we apply a force to an object with mass m, the object will accelerate at a.

Now, replacing all given values, we have

F=ma\\F=50g(20cm/s^{2} )\\F=1000\frac{gr \times cm}{s^{2} }

Therefore, the force exerted by the golf ball is F=1000\frac{gr \times cm}{s^{2} }

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Soloha48 [4]

Answer:

Solution given:

frequency[f]=x

velocity[V]=15000m/s

wave length=59m

we have

wave length=\frac{V}{f}

59m=\frac{15000}{x}

x=\frac{15000}{59}=254.Hz

frequency=254Hz

7 0
3 years ago
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If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k
Paraphin [41]
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
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6 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

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x = v't = 7.56\times 0.638 = 4.823\ m

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nadya68 [22]

Answer:

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which of the following is not a barrier to physical activity it is fear of injury I think.

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