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Lisa [10]
2 years ago
7

3B with worked out equations please

Physics
1 answer:
bekas [8.4K]2 years ago
4 0

Answer:

b) 1500N

c) to the right

Explanation:

Resultant force= 2000 -500= 1500N

Resultant force is a combination of 2 or more forces acting on an object.

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Which statement correctly describes the motion on which an Earth time interval is based?
iVinArrow [24]
The smswer is b because if there is no day there will be no month so mo year too as it takes the earth to spin 24 hours around so 24 hours is a day.
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3 years ago
How are sound waves different from light (and other electromagnetic, waves?
Blababa [14]
Sound waves travel faster
8 0
2 years ago
If a psychologist were to run an experiment measuring the effects of temperature on aggression the differing temperature would b
Yanka [14]

Answer:

(C) Independent variable

Explanation:

Independent is the variable used by the experimenter to vary a condition of interest in the experiment (temperature in this case). This measure  is varied over a sensible range to induce changes in other aspects (dependent variables; level of aggression in this case) which are measured during the experiment.

4 0
3 years ago
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

Then the relation between two different frequencies with same wavelength would be

\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}

\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

\frac{v'}{v} = \sqrt{0.96}

Finally substituting this value at the frequency beat equation we have

f_{beat} = 590(1-\sqrt{0-96})

f_{beat} = 11.92Hz

Therefore the beats per second are 11.92Hz

4 0
2 years ago
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