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3 years ago

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A uniform-density sphere whose mass is 13 kg and radius is 0.3 m

1 answer:

geniusboy [140]3 years ago

6 0

What are you trying to find?

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3 years ago

A baseball is thrown straight downward with an initial speed of 40 ft=s from the top of the Washington Monu- ment (555 ft high).

sergij07 [2.7K]

Answer:

4.53 second, 195 ft/s

Explanation:

u = 50 ft/s

h = 555 ft

g = 32 ft/s^2

Let the time taken by the ball to reach the ground is t and the velocity of the ball as it hits the ground is v.

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2} = 50^{2}+2\times 32 \times 555$

v = 195 ft /s

Thus, the ball strikes the ground with velocity 195 ft/s.

Use first equation of motion

v = u + at

195 = 50 + 32 t

32 t = 145

t = 4.53 second

Thus, the ball reach the ground in 4.53 second.

8 0

3 years ago

PHYSICS HELP<br> PLEASE HELP ITS ABOUT ATWOOD MACHINES

m_a_m_a [10]

Answer:

7.23407 $\frac{m}{s^2}$

Explanation:

(I will not include units in calculations)

I'm assuming FBD's are already drawn, so I will work from there.

Let the 2.2kg block equal $m_2$ , and the 20kg block equal $m_1$ .

Summation equation for $m_2$ : $\sum F_x=F_t_2-(F_f+F_g_x)=m_2a$ , $\sum F_y=F_n-F_g_y=0$

Summation equation for $m_1$ : $\sum F_y=F_g-F_t_1=m_1a$

Torque Summation Equation: $\sum\tau=F_t_1*r-F_t_2*r=I\alpha$

Do some plugging in with the values given: $\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha$

Replace $\alpha$ with $\frac{a}{r}$ , and cancel out the r's.

$\sum\tau=F_t_1-F_t_2=.5Ma$

This step is important: Rearrange the force summation equation to solve for each tension force.

$F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a$

Perform Substitution: $\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma$

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that $F_f=$ μ $F_n$

And based on our earlier summation equation: $F_n=F_g_y$

First, break $F_g$ into x and y components. $F_g_y=F_g\cos(\theta)$ , $F_g_x=F_g\sin(\theta)$

Perform substitution with this and the fact that $F_g=mg$ .

$\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma$

Solving for a, plugging in numbers yields an answer of 7.23407 $\frac{m}{s^2}$

6 0

3 years ago