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1 year ago

When will you say a body is in a) uniform acceleration (b) non uniform acceleration

Physics

1 answer:

s2008m [1.1K]1 year ago

7 0

Answer:

See below ~

Explanation:

Part (a) :

We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.

Part (b) :

We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.

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2 years ago

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What type of plant tissue is NOT part of the vascular bundle?

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Cortex

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2 years ago

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a microwave uses a 14 a of electricity if 120 v is run through it what is the resistance of the microwave

Airida [17]

Resistance = voltage / current.

That's. 120v / 14A = 8.57 ohms.

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3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

2 years ago

A police car with its 300-Hz siren is moving toward a warehouse at 30 m/s, intending to crash through the door. The sound bounce

Lady bird [3.3K]

Answer: The frequency heard will be f = 275.675Hz

Explanation: When an object emitting sound is moving, it occurs a phenomenon called Doppler shift or Doppler effect. What happens is that the sound gets higher when the moving object comes closer the observer and becomes lower after it passes, This change is due to the quantity of waves that passes through an area in an unit of time.

The formula to calculate the Doppler effect is as follows

f = ( $\frac{c}{c+Vs}$ ) · f₀

f is the observed frequency;

c is the speed of sound;

Vs is velocity of the source;

f₀ is the emitted frequency of source;

Substituting and calculating,

f = $\frac{340}{340+30}$ · 300

f = 275.675 Hz

Thus, the frequency heard by the police officer is 275.675Hz.

8 0

3 years ago