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Allushta [10]
3 years ago
13

Calculate the speed of an object which Travels 30 km distance in 4 hours​

Physics
1 answer:
nexus9112 [7]3 years ago
7 0

Answer:

The object is traveling 7.5 km per hour.

Explanation:

30/4 = 7.5

To check do 7.5 * 4 and you get 30

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a balloon inflated in a room at 297k has a volume of 4.00 l. the balloon is then heated to a temperature of 331 k. what is the n
Naddika [18.5K]
V2 = 4.4579 L

Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2

4.00 L = V2
———- ———
297 K 331 K

Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)

V2 = 4.4579 L

Round to significant digits if required
7 0
2 years ago
A 55 kg person falling with a velocity of 0.6
siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0
3 years ago
Garrett Hardin's "Tragedy of the Commons" describes the tendency of people to overuse publicly-owned natural resources until the
Aliun [14]

Answer:

is it bad if i keep thinking about p ussy

Explanation:

8 0
3 years ago
A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f
sergejj [24]

Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?

(ignore air resistance) (include units and correct number of significant figures)

Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888

3 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
4 years ago
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