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tankabanditka [31]
3 years ago
9

If an object is moving to the right and there is a net force acting on it to the left, what will happen to the object?

Physics
1 answer:
Ksju [112]3 years ago
7 0
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3 years ago
How much energy is required to heat 70 g of water at 20°C to boiling
choli [55]

Answer:

Q=23,430J

Explanation:

Hello,

In this case, since we compute the required energy via:

Q=mC\Delta T

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

\Delta T=100-20=80\°C

Therefore, the energy turns out:

Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J

Best regards.

3 0
3 years ago
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire
alexandr1967 [171]

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

 and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

5 0
3 years ago
What is efficiency of a machine?
Flura [38]
Mechanical efficiency is a measure of how well the machine converts the input work or energy into some useful output. It is calculated by dividing the output work by the input work. The ideal machine has mechanical efficiency equal to unity, while the real machine has mechanical efficiency less than unity
6 0
3 years ago
Read 2 more answers
a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a
Whitepunk [10]

Answer:

W = -2.76\times 10^{-9}~J

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

U = \frac{1}{2}CV^2

where C can be calculated using the plate separation and area.

C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon

Therefore, the potential energy in the first case is

U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon

In the second case:

C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J

6 0
3 years ago
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