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3 years ago

Answer fast plz ....................

Physics

1 answer:

luda_lava [24]3 years ago

8 0

Answer:

114 m/s

Explanation:

see the image below

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A child blows a leaf from rest straight up in the air. The leaf has a constant upward acceleration of magnitude 1.0\,\dfrac{\tex

Novosadov [1.4K]

Answer:

Explanation:

Given

Acceleration a = 1.0m/s²

Displacement S = 1.0m

Required

Time t taken by the leaf to displace

Using the equation of motion

S = ut+1/2at²

Substitute

1.0 = 0+1/2(1)t²

1 = t²/2

Cross multiply

t² = 2

t = ±√2

t = 1.41secs

It takes the leaf to 1.41s to displace by 1m upward

6 0

3 years ago

A 2 kg object is traveling in a circular path. What is the centripetal acceleration and centripetal force if the radius of the p

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Answer:

a = v²/R = 10²/6 = 16.7 m/s²

F = mv²/R = 2(16.7) = 33.3 N

Explanation:

3 0

3 years ago

Which principle allows us to conclude that gravity acts as the same today and will tomorrow?

Svet_ta [14]

Answer:

Gravity is a constant quantity

Explanation:

4 0

3 years ago

What is the angle o in the triangle below? (Hint: Use an inverse trigonometric<br> function.)

Margaret [11]

Answer:

Zero angle is possible geometrically in a right triangle when its length of opposite side is equal to zero. Due to zero length of opposite side, the length of hypotenuse is absolutely equal to the length of hypotenuse.

5 0

3 years ago

How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?

Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

$Q = ms\Delta T$

We know that

melting temperature of copper = 1085 degree C

Specific heat capacity of copper = 385 J/kg C

now we have

$Q = 3(385)(1085 - 83)$

$Q = 1157310 J$

$Q = 1157.3 kJ$

now in order to melt the copper we know the heat required is

$Q = mL$

here we know that

L = 205 kJ/kg

now from above formula

$Q = 3(205) kJ$

$Q = 615 kJ$

now total heat required will be

$Q = 1157.3 kJ + 615 kJ$

$Q = 1772.3 kJ$

As we know that

$1 Cal = 4.18 kJ$

now we have

$Q = \frac{1772.3}{4.18} = 430 KCal$

6 0

3 years ago