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3 years ago

6

Help me plssssssss cause I’m struggling

2 answers:

OLEGan [10]3 years ago

8 0

I think it’s c but I am not sure

krok68 [10]3 years ago

5 0

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

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What is the difference between kinetic energy and potential energy

uranmaximum [27]

Answer:

Kinetic energy is the energy of a moving body while potential energy is the energy by vutue of it's position

8 0

3 years ago

An object has a starting velocity of 20 m/s. If it accelerates at a rate of 2 m/s' for 3 seconds,

Ghella [55]

Answer:

26 m/s

69 m

Explanation:

Given:

v₀ = 20 m/s

a = 2 m/s²

t = 3 s

Find: v and Δx

v = at + v₀

v = (2 m/s²) (3 s) + 20 m/s

v = 26 m/s

Δx = v₀ t + ½ at²

Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²

Δx = 69 m

3 0

3 years ago

If a thief jumps from the tried floor of a house while holding a box on

Ksju [112]

What do you mean? it doesn't make sense

7 0

3 years ago

Read 2 more answers

What does the addition of two vectors give you?

kirill [66]

Answer:

D. Resultant Vector

Explanation:

By definition, adding 2 vectors gives a resultant vector

6 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