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3 years ago

1 point

Physics

1 answer:

Anettt [7]3 years ago

6 0

Answer:

6 hours

Explanation:

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Joules constant, 4.1868 J/cal, is an equivalence relation for 4.1868 Joules of work for 1 calorie of heat delivered to a substan

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Answer:

The final answer is $6.1 BTU = 6.5621*103 KJ$

Explanation:

$1 BTU= 252 cal\\1 BTU =252 × 4.2 kJ\\6.1 BTU = 252 × 6.1×4.2 kJ = 6562.08 kJ\\ = 6.5621 × 103 kJ\\$

Hence, $6.1 BTU = 6.5621 × 103 kJ$

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3 years ago

Jada was roller skating at 8 m/s with a total mass of 30 kg. Korbin gently pushed her and she increased her speed to 11 m/s. Fin

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Answer:

she added more force, she went down a hill with more friction.

Explanation:

either one of them

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3 years ago

1. A group of students were trying to find the greatest

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Answer:

D th

Explanation:

D B. The force applied to the ball is a balanced force.

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3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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3 years ago

Which part of the water cycle is where vapor from plants leaves the plants as they breath?

Ugo [173]

Answer:

I think it is transpiration

6 0

2 years ago

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