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Nina [5.8K]
3 years ago
10

The ice in his drink has a mass of 6 g and a volume of 20 cm3. what is the density of the ice?

Physics
1 answer:
NARA [144]3 years ago
3 0
I'm pretty sure 0.3 is the answer. Because P(density) = m/V  
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A porter can carry 40 bricks of 10 N load of each. He can carry up to 75m in 40 sec, calculate his power.​
alexira [117]

Answer:

750W

Explanation:

40×10= 400N

work done= force × distance

=400 × 75

=30000 J

Power= work done/ time

= 30000 ÷ 40

= 750 W

4 0
1 year ago
Which of the below statements are true?
andrezito [222]

Answer:

B

Explanation:

Displacement is the distance from the start point to the endpoint, displacement disregard the path taken or the amount traveled.

if you start at point A, then go to point B, and back to point A, the displacement is zero because you started and ended at the same point.

for this question, pretend you started at point A, went east 20 km to point B, and then west 8 km to point C, your displacement is 12 km. 12 km is the distance between point A and point C.

4 0
3 years ago
Why does sound travel faster in water than in air?
mars1129 [50]

Answer:

C.

Explanation:

3 0
3 years ago
Read 2 more answers
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr
Sedbober [7]

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

4 0
3 years ago
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