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3 years ago

What must occur for work to be done on an object?

Physics

2 answers:

Slav-nsk [51]3 years ago

6 0

The answer is D, The Object must move.

stiv31 [10]3 years ago

4 0

The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs

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Which statement about van der Waals forces is true?

kobusy [5.1K]

The answer is A. When the forces are weaker, they will not be able to hold the particles of the substances together; therefore, the substance will be observed as being volatile.

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4 years ago

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What is the wave speed of a 6 m wave with a frequency of 20 Hz?

andrew-mc [135]

Answer: Wave speed= frequency x wavelength

=20 x 3

=60 m/s

Explanation:

8 0

3 years ago

What is the difference between speed and velocity?

mixer [17]

Speed is the rate of change of distance with time while velocity is the rate of change of displacement with time.
Speed is a scalar quantity while velocity is a vector quantity.
Speed cannot be negative but velocity can be negative.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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3 years ago

Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

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3 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago