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2 years ago

5.

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

8 0

Answer: black

Explanation: When green light is shone on a red object, it absorbs all of the green light and not reflecting anything. Hence, it appears black.

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I will mark you as brainliest if you answer correctly

aleksklad [387]

(A) We can solve the problem by using Ohm's law, which states:
$V=IR$
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know $V=220 V$ and $R=220 \Omega$ , therefore we can rearrange Ohm's law to find the current through the device:
$I= \frac{V}{R}= \frac{220 V}{220 \Omega}=1 A$

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
$R= \rho \frac{L}{A}$
where
$\rho$ is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.

6 0

3 years ago

Calculate the gravitational potential energy of 3kg book sitting on shelf 4m high

likoan [24]

Answer:

i know it i know it pick me

Explanation:

5 0

3 years ago

Read 2 more answers

A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy asso

jasenka [17]

Answer:

$U(3)=-43J$

Explanation:

Potential energy is minus the integral of Fdx. Doing the integration yields:

$U=\int\limits {6.0x+12}\, dx$

$U=-3x^2-12x+C$

$U(0)=20J$

$U(0)=-3(0)^2+12(0)+C=20$

$C=20$

Now for x=3.0m

$U(3)=-3*(3)^2-12(3)+20$

$U(3)=-43J$

6 0

3 years ago

A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

qaws [65]

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

4 0

2 years ago

A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before fal

Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0

3 years ago

Read 2 more answers