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kkurt [141]
3 years ago
9

pls help with my with my homework, i started it but i’m stuck on the rest. if my answers that i already put are wrong pls let me

know

Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

You are 100% right for question a and you have a good start on b!

Explanation:

For (ii) of part a, you could also say that since the friction force is in the opposite direction of the pushing force, that the friction is '-60N' The negative sign just shows the direction of friction relative to the pushing force. But for the magnitude the signs are not important.

For part b,

work done equals force × distance

w = F×d

In this case, the worker is pushing with 60 newtons, and the distance is 28m.

Therefore,

w = 60N×28m

w = 1680j

(Or 1.7kj)

Great job! You got most of it!

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makkiz [27]

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6 0
3 years ago
An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s
DENIUS [597]
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
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6 0
3 years ago
What is a neoplasm? A. Another word for a malignant tumor B. A type of carcinoma that grows only in the mouth C. A benign swelli
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<h2>Answer: Any new and abnormal growth</h2>

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This abnormal mass can be benign or malignant (carcinogenic).

Therefore, the correct option is D.

3 0
3 years ago
a 250g moves eastward along a straight path at a constant velocity of 12 m per second calculate the momentum of the ball ​
lara31 [8.8K]

Explanation:

If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .

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3 years ago
Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by
OlgaM077 [116]

Explanation:

Area of ring \ 2{\pi} a d a

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Q=-\left(\pi R^{2}\right)

\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}

Note: Refer the image attached

8 0
3 years ago
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