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3 years ago

6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar

Physics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

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A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

Scrat [10]

Answer:

${ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}$

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♨ Question :

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ $\underbrace{ \sf{Required \: Answer : }}$

☄ Given :

Initial velocity ( u ) = 0
Final velocity ( v ) = 60.0 m/s
Time ( t ) = 7.50 s

☄ To find :

Acceleration ( a )

✒ We know ,

$\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) = \frac{Final velocity ( v ) - Initial velocity ( u)}{t} }}}}$

Substitute the values and solve for a.

➛ $\sf{a = \frac{60.0 - 0}{7.50}}$

➛ $\sf{a = \frac{60.0}{7.50}}$

➛ $\boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}$

---------------------------------------------------------------

✑ Additional Info :

When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
When a moving object comes in rest , in the case , final velocity ( v ) = 0
If the object is moving with uniform velocity , in the case , u = v.
If any object is thrown vertically upwards in the case , a = -g
When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

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4 0

3 years ago

Which statement best defines the relationship between work and energy?

alexdok [17]

The relationship between work and energy is that work can transfer energy between objects and cause a change in the form of energy.

<h3>What is energy?</h3>

Energy is simply defined as the ability to do work.

Energy possessed by any object or matter enables it to do work of various forms.

Energy can be transferred from one object to another. Also, energy can be transformed into various forms.

Therefore, the relationship between work and energy is that work can transfer energy between objects and cause a change in the form of energy.

Learn more about energy and work at: brainly.com/question/13881533

#SPJ1

5 0

2 years ago

Read 2 more answers

A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling

Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car $\Delta m_ {s}$ , this happens during a time interval

$[t, t + \Delta t]$ , in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider $t, m_ {c} (t) + \Delta m_ {s}$ and V.

State 2,

consider $t + \Delta t, m_ {c} (t), V + V \Delta V$

In this state is the mass of sand output, which

is composed of

$\Delta m_{s}, V + \Delta V$

In this way we define the Linear movement in x, like this:

$p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v$

$p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)$

$m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt$

In this way we proceed to obtain the Force

$F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}$

$F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}$

Since the mass of the second term becomes 0, the same term is eliminated, thus,

$F = m_ {c} (t) \frac {dv} {dt}$

$\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}$

$V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})$

3 0

3 years ago

The block accelerates down the 30 degree incline at 3m/s2. What is the coefficient of friction of these surfaces

dolphi86 [110]

Fnet = Fg sin 30 - Ff
ma = mg sin 30 - mew Fg cos 30
ma = mg sin 30 - mew mg cos 30
a = g sin 30 - mew gcos30
a - g sin 30 = - mew g cos 30
mew = -(a - g sin30)/(g cos 30)
mew = -(3m/s2 - 9.81sin30)/(9.81 cos 30)
mew = 0.22

8 0

3 years ago

Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine

Flauer [41]

Answer:

E≅1.2×10^7 N/C

Explanation:

First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!

Edit: I ended up adding it anyways ;P

Since we are observing the net electric field acting at q1, we need to use the formula: $E=k\frac{q}{r^{2} }$

And since we are observing the effects of multiple charges at once...

E=ΣE, which just means wee need to add all the observed electric fields together:

ΣE= $k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }$

Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:

q2=q3
they are the same distance from q1

These are actually really nice to have, because now we can simplify our expression to:

$E=k\frac{2q}{r^{2} }$

Now let's plug in our values and get an answer out.

E= 2(8.99×10^9)(4×10^-5)/(0.24)

Plugging all that in, I get:

E≅1.2×10^7 N/C

If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.

Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!

Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.

6 0

3 years ago