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Afina-wow [57]
3 years ago
14

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200 WW electric im

mersion heater in 0.400 kgkg of water.
Physics
1 answer:
ivolga24 [154]3 years ago
7 0

Answer:

The heat is 115478.4 J.

Explanation:

Given that,

Mass of water = 0.400 kg

Power = 200 W

Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?

We need to calculate the heat

Using formula of heat

Q=mc\Delta T

Where, m = mass of water

c = specific heat

Put the value into the formula

Q=400\times4.184\times(89-20)

Q=115478.4\ J

Hence, The heat is 115478.4 J.

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During the push-up, the hips should never hit the ground and should move 1 point
Monica [59]
Hahahahahahahahahha true
4 0
3 years ago
If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the
erastova [34]

Either 175 N or 157 N depending upon how the value of 48° was measured from.    
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.    
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
U = Useful force  
F = Force applied    
So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
117 N/0.669130606 = F  
174.8537563 N = F    
So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
U/cos(42) = F  
117/0.743144825 = F  
157.4390294 = F    
Or 157 Newtons is required for this case.
6 0
2 years ago
How much work is done when a porcelain vase with mass of 2.5 kg is lifted up 3 m to a shelf? (HINT- Weight can be used as Force
alexdok [17]
Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
Work = 7.5 J
J=Jules (Jules is the unit uses to calculate work)
8 0
2 years ago
A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l
son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Using formula of focal length for glass ethyl alcohol system

\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})

Divided equation (II) by (I)

\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}

Where, n_{g} = refractive index of glass

n_{a} = refractive index of air

n_{ethyl} = refractive index of ethyl

Put the value into the formula

\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}

\dfrac{f'}{11.5}=\dfrac{25}{7}

f'=\dfrac{25}{7}\times11.5

f'=41.07\ cm

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0
3 years ago
If you push on a wall with a force of 200 N, with what force does the wall push back?
PSYCHO15rus [73]
200 N, that is if the force is balanced and the wall doesn't move
8 0
2 years ago
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