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Mkey [24]
3 years ago
10

Suppose you take a trip that covers 180 km and takes 3 hours to make.

Physics
1 answer:
Allushta [10]3 years ago
5 0

Answer:

v = 16,66 m/s

Explanation:

To obtain the velocity of the train we must use the velocity formula for a uniform line movement:

v = x/t

Where x is the space and t is time.

replacing given values:

v = 180 km / 3 h

v = 60 km/h

to pass this value to international units:

v = 60 / 3,6 m/s

v = 16,66 m/s

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A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
Hey can anyone help me out in dis pls!
patriot [66]

Answer:

D

Explanation:

3 0
3 years ago
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This type of liquid can be used as an electric tester Is that a good conducted of electricity?
Helga [31]

Answer:

Yes

Explanation:

Testing electricity is a good conductor of electricity.

7 0
2 years ago
In the case of mechanical waves, what causes or determines the speed of the waves, the frequency of the waves, and the wavelengt
Levart [38]
The rate in witch ditermans the speed or vibration of the movment under the waves witch couses vibrational freequencys to be disrupted.
3 0
3 years ago
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Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by
OlgaM077 [116]

Explanation:

Area of ring \ 2{\pi} a d a

Charge of on ring d q=-(\ 2{\pi} a d a)

Charge on disk

Q=-\left(\pi R^{2}\right)

\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}

Note: Refer the image attached

8 0
3 years ago
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