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Elodia [21]
3 years ago
10

The waves that heat a cup of water in the microwave are an example of electromagnetic waves. True or False

Physics
1 answer:
Oksana_A [137]3 years ago
5 0

That's true.

They're radio waves, at the frequency of 2.45 GHz (in all microwave appliances manufactured in the US).

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A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'
Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom v=4.22 m/s

Diameter of loop d=28.6 cm

As Energy is conserved therefore Energy at top is equal to energy at bottom

E_T=E_B

\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}  ,where v_0 is the velocity at bottom

\frac{v^2}{2}+gh=\frac{v_0^2}{2}

v_0^2=v^2+2gh

v^2=v_0^2-2gh

v=\sqrt{v_0^2-2gh}

v=\sqrt{4.22^2-2\times 9.8\times 0.286}

v=\sqrt{17.8084-5.6056}

v=3.49 m/s

                       

7 0
3 years ago
Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72 s
Sladkaya [172]
1.5 m/s toward shore 

8 0
3 years ago
Jack travels 80 km in 50 minutes. What is his speed in m/s?
Pie

Answer:

1.6 m/s?

Explanation:

80 ÷ 50 = 1.6 m/s

djdjdjksjsj

7 0
3 years ago
A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill
Stolb23 [73]

Answer:

a.) the speed at the bottom is greater for the steeperhill

Explanation:

since the energy at the bottom of the steeper hilis greater

mgh =\frac{1}{mv^2}

As we can see from above that v is higher when h ishigher.

8 0
3 years ago
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
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