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bagirrra123 [75]
4 years ago
12

When you do positive work on an object, you ______

Physics
2 answers:
guajiro [1.7K]4 years ago
4 0

C. increase the object's energy

gtnhenbr [62]4 years ago
3 0

when you do positive work on an object, you ______  

A. decrease the object's energy

<u>B. keep the object's energy the same </u>

C. increase the object's energy

D. may increase or decrease the object's energy

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What is the wavelength of an earthquake wave if it has a speed of 3 km/s and a frequency of 6 Hz?
disa [49]
Wavelength = c/f.
Wavelength =0.5km
8 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
i cant figure out how to delete a question on this website. so i am writing this until i figure it out.
sergij07 [2.7K]

Answer:

I'm guessing it would remain a mystery to us all

5 0
3 years ago
How many watts are used when 6000 joules of energy are consumed in 5 hours?
Ray Of Light [21]
Power = energy / time
Time should be in seconds
5 hours = 5 × 60 minutes = 5 × 60 ×60 seconds = 18000 s
Power = 6000/18000 = 0.333 W

Hope it helped!
7 0
3 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

4 0
3 years ago
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