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4 years ago

Which two statements are true about a system

Physics

1 answer:

kykrilka [37]4 years ago

5 0

First one is what are the options
second one is the system in this question is no use

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Diana hangs a lead cylinder with a force of 12 N from a model bridge.

vova2212 [387]

The answer would be A and I’m pretty sure of thta

7 0

3 years ago

A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during

nikdorinn [45]

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

S = ut + ½ at²

Substituting the given values,

S = 30 x 10 + ½ x 10 x 10²

S = 800 m

Hence, the distance covered by the body is, S = 800 m

5 0

3 years ago

6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces

Vesna [10]

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket $V=15\ gal\approx 56.78\ L$

time to fill it $t=50\ s$

Volume flow rate

$\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s$

The inner diameter of the hose $D=1.5\ cm$

diameter of the nozzle exit $d=0.8\ cm$

we can volume flow rate as

$\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s$

Mass flow rate

$\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s$

5 0

3 years ago

After 60 days, 100g of a certain element has decayed to only 12.5g.

vagabundo [1.1K]

Answer:

8 days

Explanation:

3 0

3 years ago

A student went to a hill station early in the morning, he could hear the echo of his clap after 0.1 second. When he went to the

Dahasolnce [82]

Answer:

See the explanation.

Explanation:

The speeds of sound depend upon the temperature of the medium. As the temperature increased in the afternoon, the speed of sound increases. Then the time taken by reflected sound will be less than 0.1 sec in the afternoon. And to hear an echo the time gap between an original sound and reflected sound must be at least 0.1 sec. That is why the student could not hear the echo at all in the afternoon.

7 0

3 years ago