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Alja [10]
3 years ago
10

A bowler once measured that she can throw the bowling ball with a speed of 15miles/hour.If it takes 3 seconds from the ball to t

ravel from the beginning of the lane down to the pins,how far is this distance in feet?
Physics
1 answer:
Vinvika [58]3 years ago
8 0

The distance traveled by the ball in feet is 66 ft.

The given parameters;

  • speed of the ball, u = 15 miles/hour
  • time of motion of the ball, t = 3 seconds

1 mile = 5280 feet

1 hour = 3600 s

The speed of the ball in feet per second is calculated as follows;

v = \frac{15 \ miles}{hour} \times \frac{5280 \ ft}{1 \ mile} \times \frac{1 \ hour}{3600 \ s} = 22 \ ft/s

The distance traveled by the ball in feet is calculated as follows;

Distance = speed x time

Distance = 22 ft/s x 3 s

Distance = 66 ft

Thus, the distance traveled by the ball in feet is 66 ft.

Learn more here: brainly.com/question/13722298

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An electromagnet is made from a coil of wire which acts as a magnet when an electric current passes through it. Often an electromagnet is wrapped around a core of ferromagnetic material like steel, which enhances the magnetic field produced by the coil.

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In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
Nataliya [291]

Answer:

A) mr = 100 kg

B) Fr = 210N

C) Ft = -199.5N

Explanation:

By conservation of the momentum:

mt*Vo = (mr + mt) * Vf  Solving for mr:

mr = mt*Vo / Vf - mt = 100 kg

The average force on the receiver:

mr *(Vf - 0) = Fr * Δt    Solving for Fr:

Fr = 210 N

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mt * (Vf - Vo) = Ft * Δt    Solving for Ft:

Ft = -199.5 N

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8 0
4 years ago
Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient
Luba_88 [7]

Answer:

a. a = 6.41 m/s^2

b. T = -0.81 N

Explanation:

Given,

  • weight of the lighter block = w_1\ =\ 3.0\ N
  • weight of the heavier block = w_2\ =\ 7.0\ N
  • inclination angle = \theta\ =\ 30^o
  • coefficient of kinetic friction between the lighter block and the surface = \mu_1\ =\ 0.13
  • coefficient of kinetic friction between the heavier block and the surface = \mu_2\ =\ 0.31
  • friction force on the lighter block = f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta
  • friction force on the heavier block = f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

From the f.b.d. of the heavier block,

w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)

From eqn (1) and (2), we get,

w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\

\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\

\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.

part (b)

From the eqn (2), we get,T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N

3 0
3 years ago
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