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Molodets [167]
3 years ago
8

Using the projectiles equations for y and x at what second does the ball have a vertical velocity of -39.2 m/s^2

Physics
1 answer:
vovangra [49]3 years ago
8 0

Answer:

<h2>a) t = 4</h2>

Explanation:

In projectile the time taken by an object to reach its maximum height is the maximum time (tmax).

tmax = Usin\theta/g

U = velocity of the ball

\theta = angle of projection = 90° (vertical velocity)

g = acceleration due to gravity = 9.8m/s²

tmax = Usin90°/9.8

tmax = U/9.8

T the maximum height U = Uy =  39.2 m/s²

tmax = 39.2/9.8

tmax = 4secs

To calculate the time at which the ball will have an horizontal velocity, we will use the same formula but \theta will be 0° in this case

If Ux = Ucos\theta

8 = Ucos0

U = 8m/s

Since t = Usin\theta/g

t = 8sin0/9.8

t = 0sec

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Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are gi
creativ13 [48]

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)\neq0

d) 2+2+(-4)=0

e) 2+2+(-2)\neq0

f) 2+2+(-3) \neq0; 2+(-2)+3\neq0

g) 2+2+(-5)\neq0; 2+(-2)+5\neq0

h)200 + 200 +(-5)\neq0; 200+(-200)+5\neq0

6 0
3 years ago
A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be near
alina1380 [7]

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

  • That the temperature on the day is 35 °C
  • That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

3 0
3 years ago
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