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3 years ago

Suppose that a pendulum has a period of

Physics

1 answer:

Yakvenalex [24]3 years ago

4 0

Explanation:

period of pendulum = time taken for 1 oscillation = time taken for 1 complete back and forth vibration

q1 ans is given in question its 1.5 sec

q2 ans is 1.5 sec longer than 1 sec period

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Jordan wants to know the difference between using a 60-W and 100-W lightbulb in her lamp. She calculates the energy it would tak

Andreas93 [3]

Answer:

The difference in the amount of energy transferred by the two bulbs is 1200 J.

Explanation:

The energy transferred by the two lightbulbs can be calculated with the given equation:

$E = P*t$

Where P is the power and t is the time

For the 60 W lightbulb:

$E_{60} = P*t = 60 W*30 s = 1800 J$

For the 100 W lightbulb:

$E_{100} = P*t = 100 W*30 s = 3000 J$

Hence, the difference in the amount of energy transferred is:

$E_{t} = E_{100} - E_{60} = 3000 J - 1800 J = 1200 J$

Therefore, the difference in the amount of energy transferred by the two bulbs is 1200 J.

I hope it helps you!

4 0

3 years ago

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

$h_{A}$ = 100 m

$v_{A}$ = $v_{A}$

$h_{B}$ = 150 m

$v_{B}$ = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed $v_{A}$

final speed $v_{B}$ = 0

thus,

$v_{B}$ ² = $v_{A}$ ² - 2 g h

0 = $v_{A}$ ² - 2 g h

$v_{A}$ ² = 2 g h

$v_{A}$ = √2 g h

= √2 (9.8) (50)

= 31.3 m/s

8 0

3 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$