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3 years ago

A train travelled 500 meters in 25 seconds

Physics

2 answers:

KIM [24]3 years ago

7 0

$\huge\mathfrak\red {Answer:}$

<h2>Speed = Distance/Time</h2>

If a train travelled 500 meters in 25 seconds then,

Speed = 500m/25sec

$\mathfrak\purple {Hope\: this\: helps\: you...}$

Radda [10]3 years ago

7 0

speed=500m/25sec

20m/sec

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a

iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

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3 years ago

A pendulum is observed to complete 20 full cycles in 60 seconds. Determine the period and the frequency of the pendulum.

Delvig [45]

i hope i have been useful buddy.

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3 years ago

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A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

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3 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

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3 years ago

Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)

PilotLPTM [1.2K]

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

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1 year ago