All categories

1 year ago

How is force useful as well as harmful.Comment with the statement.

1 answer:

vladimir1956 [14]1 year ago

7 0

No, we can't say force is useful or harmful. All else being equal, force is either not harmful or useful depending on the body that is applying it and where it is applying it.

Given how it affects motion, force is a crucial idea. It is a relationship that, in the absence of an opposing force, modifies an object's motion. However, a push or a pull that any object feels is the simplest definition of force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction.

How force is useful

A body at rest can move with enough force.A body in motion may be slowed down or stopped by it.It has the power to quicken the pace of an object in motion.Along with its shape and size, it can also alter the direction of a moving body.

How force is harmful

Force has the power to alter an object's state of motion.
Moving objects can shift direction due to force.
Moving things' speeds can be increased by force.
Moving items can become slower due to force.
Force has the power to alter an object's shape.

Learn more about force here

brainly.com/question/14362949

#SPJ9

You might be interested in

t or f if an object has a canstant acceleration it must ve changing velocity at the same rate over over again

Angelina_Jolie [31]

The answer to your question is True

7 0

3 years ago

a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second

Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
1000
k
H
z
=
1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = 30.16 x 10^30 photon/s.

1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.

Learn more about photon on:
brainly.com/question/20912241

#SPJ4

6 0

11 months ago

What radiation lies at frequencies just below the frequencies of visible light?

vlabodo [156]

Infrared radiation also called as infrared light was discovered by Sir William Herschel in 1800.It is an electronic magnetic radiation with longer wavelengths than those of a visible light.It involves waves rather than particles. It lies at frequencies just below the frequencies of visible light.

3 0

3 years ago

Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l

Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1 (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2 (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2) (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5 (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2 (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

$R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067} \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22$

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

= 0.5 - 0.9931 × 0.340

= 0.5 - 0.338

= 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0

2 years ago

What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

dimulka [17.4K]

Answer:??

Explanation:

5 0

2 years ago

How is force useful as well as harmful.Comment with the statement.​

How is force useful as well as harmful.Comment with the statement.