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Reptile [31]
3 years ago
6

Give your answer with solution(Easy Question)

Physics
1 answer:
Furkat [3]3 years ago
5 0

Volume of any rectangular prism is (length) x (width) x (height).

The box in the picture is definitely a rectangular prism, so its volume is

(length) x (width) x (height) .

That's (4.0 cm) x (6.0 cm) x (2.0 cm) = 48 cm³ .

So the volume of the box is 48 cm³ fer sher, but that's NOT the preferred answer to this question.  If you take your time and read all the fine print carefully, you'll see that this sneaky question wants the answer in units of meter³ .  In order to produce the answer that it's fishing for, YOU have to be sneaky too.  You have to convert units, either before you start doing the calculating, or else convert the units of your result before you choose the answer from the list.

Before you calculate: Convert the given dimensions to (meters).  Then

Volume = (0.04 m) x (0.06 m) x (0.02 m)

Volume = 0.000048 m³

<em>Volume = 4.8 x 10⁻⁵ m³</em>


After calculating but before picking the answer:

Volume = (4.0 cm) x (6.0 cm) x (2.0 cm)

Volume = (48 cm³)

Convert:

Volume = (48 cm³) x (1 meter / 100 cm)³

Volume = (48 cm³) x (1 meter³ / 10⁶ cm³)

Volume = (48 cm³ meter³ / 10⁶ cm³)

Volume = 48 x 10⁻⁶ m³

or

<em>Volume = 4.8 x 10⁻⁵ m³</em>

This is exactly the same number as 48 cm³, but this is the way they want to see it, and 48 cm³ isn't.  So you have to select choice - <em>D .</em>


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. How much work in joules is done by a person who uses a force of 25 N to move a desk 3.0 m?
sergejj [24]
We Know, W = F * s
here, F = 25 N
s = 3 m

Substitute it into the expression,
W = 25 * 3
W = 75 Joule

So, your final answer is 75 J

Hope this helps!

7 0
3 years ago
The kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.
eimsori [14]
E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J
8 0
3 years ago
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

6 0
3 years ago
if a piece of sea floor has moved 50 km in 5 million years what is the yearly rate of sea-floor motion?
BartSMP [9]
<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
6 0
3 years ago
Read 2 more answers
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

Finally:

U_{f}=14.1 m/s

8 0
3 years ago
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