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3 years ago

Which types of basketball injuries do you believe the warm-up could prevent? Explain

Physics

1 answer:

inn [45]3 years ago

7 0

Answer:

pulling a muscle. putting too much strain on a muscle causing a tear.

Explanation:

Doing warm ups help your muscles adjust to the full extent of a basketball game. Warming up is like stretching, helps your muscles adjust before going full out on the court.

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It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is

KiRa [710]

Answer: 4 ft/s

Explanation:

Given

height of man $=4 ft$

speed of person $v=8 ft/s$

height if street light $=12 ft$

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

$\frac{4}{12}=\frac{y}{x+y}$

$\frac{1}{3}=\frac{y}{x+y}$

$x+y=3y$

$x=2y$

differentiate above Equation w.r.t time we get

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{8}{2}=4 ft/s$

6 0

3 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

3 years ago

A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

LenaWriter [7]

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

7 0

3 years ago

The manufacturer of a bulletproof vest wants the vest to be able to stop a bullet with a mass of 0.4 kg and a velocity of 1800 m

11111nata11111 [884]

We know that momentum = mass times velocity
So a. 720 kgm/s

7 0

3 years ago

8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

pantera1 [17]

(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that

0² - v² = 2 a d → a = - v² / (2d)

Then the car comes to rest over a distance of

d = - v² / (2a)

Doubling the starting speed gives

- (2v)² / (2a) = - 4v² / (2a) = 4d

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration a such that

0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass m such that

60 J = m g h

where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives

m = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0

3 years ago