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3 years ago

The rate at which energy is transferred is called a. joules. b. power. c. work. d. time.

Physics

2 answers:

Marrrta [24]3 years ago

7 0

B. Power
I hope this helps you, have a great day!

Molodets [167]3 years ago

6 0

B. Power.
___________

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A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

q = 0.0000249 C

$q=2.49\times 10^{-5}\ C$

Hence, this is the required solution.

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3 years ago

Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitud

Gekata [30.6K]

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

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3 years ago

You are pushing a 30-kg block on a rough floor in a direction that is parallel to the floor. The block moves with a uniform 2 m/

kykrilka [37]

Answer:dd

Explanation:

I dunno

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2 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

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3 years ago

How to atoms behave in non-magnetic items?

Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

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3 years ago