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raketka [301]
3 years ago
12

How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?

Physics
1 answer:
AleksandrR [38]3 years ago
7 0

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

WD= 45*6

WD=270Joues

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Use of electromagnetic because it moves very faster than others for example  xrays theynar very very slow so that not It it is d.
4 0
3 years ago
A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict
lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

a=0

This means that the net force is also zero:

\sum F=0

The net force on the block is given by the applied force, forward, and the frictional force, backward:

\sum F = F_a-F_f=0

where

F_a=39.4 N is the applied force

F_f is the frictional force

Therefore, solving for F_f,

F_f=F_a=39.4 N

Learn more about friction:

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8 0
3 years ago
I NEED THE RIGHT ANSWER ASAP NO LINKS !!!<br> This is a Science question
sdas [7]
3rd number line -5- -4 is -9
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8 0
3 years ago
Read 2 more answers
A current of 310 amps is flowing through a copper wire with a resistance of 30 ohms. What is the voltage?
torisob [31]

Answer:

the voltage is 930V

Explanation:

U=I*R

U=310*30=930V

8 0
3 years ago
A capacitance C and an inductance L are operated at the same angular frequency.
Levart [38]

A) \omega = \frac{1}{\sqrt{LC}}

The magnitude of the capacitive reactance is given by

X_C = \frac{1}{\omega C}

where

\omega is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

X_L = \omega L

where L is the inductance.

Since we want the two reactances to be equal, we have

X_C = X_L

So we find

\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}

B) 7449 rad/s

In this case, we have

L=5.30 mH = 5.3\cdot 10^{-3}H is the inductance

C= 3.40 \mu F= 3.40 \cdot 10^{-6}F is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s

C) 39.5 \Omega

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega

- Inductive reactance:

X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega

7 0
3 years ago
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