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iVinArrow [24]
4 years ago
9

At a horse race, all horses started from rest. As soon as the gates were open the horses started to trot. After 300s, Blacky was

accelerating at 3m/s2 . What is his final velocity after 300s?
Physics
1 answer:
Bas_tet [7]4 years ago
4 0
Acceleration<span> is a vector quantity that is defined as the rate at which an object changes its velocity. An object is </span>accelerating<span> if it is changing its velocity. It can be calculated by the expression:

a = v2 - v1 / t

From the given in the problem, we can solve for v2, the final velocity:

3 = v2 - 0 / 300
v2 = 900 m/s</span>
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Goryan [66]
Yeah that’s is correct
5 0
3 years ago
A gas is taken in a vessel which is then sealed. The gas is then heated. How will the
yanalaym [24]

Answer:

D The temperature, pressure and volume will all increase

3 0
4 years ago
Read 2 more answers
A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

8 0
3 years ago
Based on the graph of kinetic energy given (gray curve in the graphing window), sketch a graph of the baseball's gravitational p
chubhunter [2.5K]

1) Physical principles:

a) Total mechanical energy = kinetic energy + potential energy.

b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)

2) Notation:

a) Total mechanical energy: ME.

b) Kinetic energy: KE

c) Gravitational potential energy: PE

∴ ME = KE + PE = constant

3) Solution:

a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.

b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0

c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).

d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.

At that point, KE = 0, and PE = ME.

That is the point of highest altitude and where the speed is zero.

d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.

Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.

The PE may be sketched on the same graph along with the KE and the ME.

The graph is attached. The red line is the ME and the blue line is the PE.

Note that at any point in the graph PE + KE = ME.

5 0
4 years ago
Read 2 more answers
Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not
d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},

where  I_0(\text{at }1000\text{ Hz}) is the hearing threshold at 1000 Hz.

Therefore we have the following

0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}

I_1 is the threshold at 1000Hz and I_2 is the threshold at 100Hz.

By exponentiating we have

10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.

Now dividing these two equations we get

\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

7 0
3 years ago
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