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4 years ago

14

answer this question: When you can not see what is taking place, but other senses indicate occurrences. This is called _________

__

1 answer:

Phoenix [80]4 years ago

5 0

This is called an indirect observation.

You might be interested in

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Lemur [1.5K]

Answer:

a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

λ = x

λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

T / 2 = t

T = 2t

period and frequency are related

f = 1 / T

f = 1 / 2t

we substitute

v = x / 2t

v = 5.50 / 2 3

v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

2A = y

A = y / 2

A = 0.700 / 2

A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

v = 0.9167 m / s

d) the amplitude is

A = 0.500 / 2

A = 0.250 m

4 0

3 years ago

If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the

KiRa [710]

Answer:

option (c) is true.

Explanation:

According to the Newton's gravitation law, the force of gravitation acts between the two bodies are same in magnitude but the direction is opposite. It obeys the newton's third law of motion.

So, F1 is the force exerted by the earth on a satellite is equal to the force F2 exerted by the satellite on the earth.

3 0

4 years ago

An air bubble released by a deep-water diver, 115 m below the surface of a lake, has a volume of 1.60 cm3. The surface of the la

Kryger [21]

Answer:

The value is $V_2 = 1.9396 *10^{-5} \ m^3$

Explanation:

From the question we are told that

The depth at which the bubble is released is $h = 115 \ m$

The volume of the air bubble is $V = 1.60 cm^3 = 1.60 *10^{-6} \ m^3$

Generally from the ideal gas law

$PV = nRT$

Given that n , R , T are constant we have that

$PV = constant$

So

$P_1 V_1=P_2 V_2$

Here $P_1$ is the pressure of the bubble at the depth where it is released which i mathematically represented as

$P_1 = P_a + P$

Here $P_a$ is the atmospheric pressure with value $P_a = 101325 \ Pa$

and $P$ is the pressure due to the depth which is mathematically represented as

$P = \rho * g * h$

So

$P = 1000 * 9.8*115$

=> $P = 1127000\ Pa$

Here $\rho$ is the density of pure water with value $\rho = 1000 \ kg/m^3$

$g = 9.8 \ m/s^2$

$V_1$ is the volume of the bubble at the depth where it is released

$P_2$ is the pressure of the bubble at the surface which is equivalent to the atmospheric temperature

$V_2$ is the volume of the bubble at the surface

So

$V_2 = \frac{ P_1 * V_1}{ P_2}$

=> $V_2 = \frac{(Pa+ P) * V_1}{P_a}$

=> $V_2 = \frac{101325 + 1127000 * (1.60 *10^{-6})}{ 101325 }$

=> $V_2 = 1.9396 *10^{-5} \ m^3$

4 0

3 years ago

The rotor in a certain electric motor is a flat, rectangular coil with 80 turns of wire and dimensions 2.50 cm by 4.00 cm . The

andreyandreev [35.5K]

The average power of the motor is 0.153 Watt

To find the average power, the given values are,

No .of turns of the wire = 80

Dimensions is given as, 2.50 cm by 4.00 cm

Magnetic field = 0.800 T.

current = 10 mA.

Angular speed = 3.60 ×10³ rev/min

What is average power?

Average power is defined as the ratio of total work done by the body to the total time taken by the body.

Average power P avg = W / ΔT Watt

W - Work in One revolution

For first one half revolution,

W = 2 NIAB

= 2× 80×10×10⁻³×0.0250×0.040×0.800× Sin 90°

= 2 × 6.4 × 10 ⁻⁴

= 1.2 × 10⁻³ J

For full revolution,

W = 2 × 1.2 × 10⁻³ J

= 2.56 × 10⁻³ J

Calculating power,

Time for one revolution is Δt = 60/3600 = 1/60s

Average power = W / Δt

= 2.56 x 10⁻³ / (1/60)

= 0.153 Watt.

The average power of the motor is 0.153 Watt.

Learn more about Average power,

brainly.com/question/27873045

#SPJ2

3 0

2 years ago

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

4 years ago