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4 years ago

Help please

Physics

2 answers:

elena-14-01-66 [18.8K]4 years ago

7 0

at a particular instant. so C.

Good luck

Talja [164]4 years ago

6 0

Instantaneous speed is measured C. at a particular instant.
Instantaneous refers to one specific moment in time, which is answer C.

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an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo

Otrada [13]

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

$F = ma$

$9000 = 6000 a$

$a = 1.5 m/s^2$

now for finding minimum stopping distance of the truck

$v_f^2 - v_i^2 = 2a d$

$0^2 - 15^2 = 2(-1.5) d$

$d = 75 m$

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4 years ago

A container of an ideal gas at 1 atm is compressed to 1/3 its volume, with the temperature held constant, what is its final pres

Sati [7]

Answer:

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Explanation:

hindi kasi magaling sa physic

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2 years ago

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A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

alexandr402 [8]

Answer:

$v = 12.57 m/s$

Explanation:

As we know that the radius of the circular motion is given as

$R = 50.00 cm$

time period of the motion is given as

$T = 0.2500 s$

now we know that it is moving with uniform speed

so it is given as

$v = \frac{2\pi R}{T}$

now plug in all data

$v = \frac{2\pi(0.50)}{0.25}$

$v = 12.57 m/s$

3 0

4 years ago

You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is

Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

Weight in air

W (air) = mg, where g is the gravitational acceleration

Weight with submerged with only one mass

m(s)g + Fb = mg + m(b)g, consider this to be equation 1

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2

equation 1 – equation 2 would give us

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

Substituting V into the above equation we get

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0

3 years ago

When the bug is stationary and creating waves, how does the frequency of the wave some distance away from the bug compare with t

Brrunno [24]

Answer:

The frequency is the same

Explanation:

When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.

This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.

In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency $f$ : as a consequence, the frequency $f'$ of the waves produced by the bug will be equal to the frequency of vibration of the bug:

$f'=f$ .

3 0

3 years ago