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My name is Ann [436]
3 years ago
12

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

hat is its velocity as it leaves his hand?
Physics
1 answer:
Aloiza [94]3 years ago
4 0

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

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A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do
sweet-ann [11.9K]

\blue{\bold{\underline{\underline{Answer:}}}}

  • \green{\tt{Work\:done=24\:J}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

\green{\underline{\bold{Given :}}} \\  \tt: \implies Constant \: force(F) = 8 \: N \\  \\ \tt: \implies Displacement(s) = 3 \: m \\  \\ \red{\underline{\bold{To \: Find : }}} \\  \tt:  \implies Work \: Done(W.D) = ?

• <u>According to given question</u> :

\green{ \star} \tt \:  \theta \:  = 0 \degree \:  \:  \:  \: (Angle \: between \: force \: and \: displacement) \\  \\  \bold{As \: we \: know \: that} \\  \tt:  \implies Work \: Done = FS \: cos  \: \theta \\  \\  \tt:  \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\  \\ \green{ \circ} \tt \: cos  \: 0 \degree = 1  \\  \\  \tt:  \implies Work \: Done =24 \times 1 \\  \\   \green{\tt:  \implies Work \: Done =24 \: J}

5 0
3 years ago
I WILL GIVE BRAINLIEST....<br>Determine the value of F...​
aleksley [76]
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.

Or it could be 15 if both top and bottom are 25. 10+15= 25.
6 0
3 years ago
a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi
alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, u=60.0\ mm/s

Angle of projection is, \theta=30.0\°

Time taken to land on the hill is, t=4\ s

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0
3 years ago
an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the el
ololo11 [35]

Answer:

1.85c

Explanation:

a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c

5 0
2 years ago
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