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2 years ago

A stationary rock on a hill has

Physics

2 answers:

Mekhanik [1.2K]2 years ago

5 0

Answer: potential energy but no kinetic energy

Explanation:

Since the rock is stationary, velocity is zero, therefore no kinetic energy,but there's potential energy because the rock is at rest,

Tresset [83]2 years ago

3 0

Answer:

C. Potential energy but no kinetic energy

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A circuit has resistors R 1 = 468 Ω and R 2 = 125 Ω R1=468 Ω and R2=125 Ω , and two batteries V 1 = 12.0 V and V 2 = 3.00 V V1=1

Greeley [361]

Answer:

0.0192A

Explanation:

Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:

Step 1

VR = V2

VR = 3.00v

Step 2

Calculating the current through the resistance R as below,

IR = V1 - V2 /R1

IR = 12 - 3 /468

IR =0.0192A

7 0

3 years ago

A machine has mechanical advantage 2.What does that mean

EleoNora [17]

Answer:

Mechanical advantage is the measure of the force amplification achieved by using a tool , mechanical device or machine . The machine preserve the input power and supply trade off force. against movement to obtain a desired amplification in the output force .

5 0

2 years ago

Read 2 more answers

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,

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