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1 year ago

7

A motorcycle traveling at 25 m/s accelerates at a rate of 6 m/s2 for 4 seconds. What is the final speed of the motorcycle in m/s

?

1 answer:

givi [52]1 year ago

8 0

Initial velocity = Vo= 25 m/s

Final velocity = V = x

Acceleration= a = 6 m/s^2

time= t = 4 seconds

Appy the equation:

V = Vo + at

Replacing:

V = 25 + 6(4) = 25 + 24 = 49 m/s

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Michaela is babysitting little Tommy Turner who is known to be terror . Michaela answered the phone when Tommy's parents called

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Answer: the ballon will explode.

Explanation:

The air in the ballon will increase its temperature and, according to the laws of behavior of the gases, it will expand until it explode.

You can assume ideal gas behavior and use PV = nRT, where you can see that the volume V and P are proportional to the temperature T, so with the increase in temperature the ballon will increase its volume, until the material of the ballon can not expand more, when the pressure will increase and the material will fail.

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49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner

faust18 [17]

Answer:

$\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}$

Explanation:

For this interesting problem, we use the definition of centripetal acceleration

a = v² / r

angular and linear velocity are related

v = w r

we substitute

a = w² r

the rectangular body rotates at an angular velocity w

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A

the distance OB = L₂

the distance AB = L₁

the sides of the rectangle

It is indicated that the acceleration in in A and B are related

$a_A = n \ a_B$

we substitute the value of the acceleration

w² r_A = n r_B

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3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

<u>Now the time required for supply of 185370 J:</u>

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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Answer:

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Explanation:

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3 years ago

You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin

Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

W₀ = G m M / Re²

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r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

r = Re + 250

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The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

W = G m M / Re² (1+ 250 / Re)²

(1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

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(1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

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W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

W = Wo (1 - 7.85 10⁻⁵)

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3 years ago

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