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AnnyKZ [126]
3 years ago
12

A 500-kilogram sports car accelerates uniformly from rest, reaching a speed of 30 meters per second in 6 seconds. During the 6 s

econds, the car has traveled a distance of:_______
(A) 15 m
(B) 30 m
(C) 60 m
(D) 90 m
(E) 180 m
Physics
1 answer:
AlekseyPX3 years ago
4 0

Answer:

90 m

Explanation:

Given that

Mass of the car, m = 500 kg

Initial speed of the car, u = 0 m/s

Final speed of the car, v = 30 m/s

Time taken to travel, t = 6 s

average speed v = (30 + 0) / 2 = 15

We are going to be using the most very basic equations to solve this.

speed can be defined as the ratio of distance with respect to time.

v = d/t, Where

v = speed.

d = distance travelled and

t = time taken

if we invert this, we can get that

d = v.t

distance is multiplication of speed and time. We will be using the average speed we calculated above, so

Distance travelled in 6 s,

d = 15 * 6

d = 90 m.

therefore, the distance travelled by the car in this 6 s is 90 m

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A rigid container holds 0.30g of hydrogen gas.
Strike441 [17]

Answer:

Part A:    \mathbf{Q =94 \ J} to two significant figures

Part B:    \mathbf{Q =160  \ J} to two significant figures

Part C:    \mathbf{Q =220  \ J} to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = C_v=\dfrac{3}{2}R

For Part A:

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K

Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)

Q=93.5325 \ J

\mathbf{Q =94 \ J} to two significant figures

Part B. For hot temperature, C_v=\dfrac{5}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K

Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)

Q=155.8875 \ J

\mathbf{Q =160  \ J} to two significant figures

Part C. For an extremely hot temperature, C_v=\dfrac{7}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K

Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)

Q=218.2425 \ J

\mathbf{Q =220  \ J} to two significant figures

6 0
3 years ago
I need help in this pls i really need a answer
IrinaK [193]

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\  = 30.0 + 10.0 + 20.0 \\  = 60.0 \: km

Total time is 3.0 hours

but:

average \: speed =  \frac{total \: distance}{total \: time}  \\

In terms of distance.

substitute:

average \: speed =  \frac{60.0}{3.0}  \\  \\  = 20 \:  {kmh}^{ - 1}

displacement = ( - 30.0) + 10 .0+ 20.0 \\  = 0

In terms of displacement:

speed =  \frac{0}{3}  \\  \\  = 0 \:  {kmh}^{ - 1}

3 0
3 years ago
What type of interaction allows the sidewalk to heat up and cook the egg? 
Bogdan [553]

Answer:

Adsorption

Explanation:

Sidewalk cooking of egg is very popular in America. In summer people release fireworks in night sky and cook egg on the concrete sidewalk to check the level of temperature. When sunlight fall on the sidewalk most of the light is reflected back but some darker material adsorbs some photon, and when these photons are transferred to egg molecules it causes vibration among them and produce heat and cook egg.

7 0
3 years ago
Read 2 more answers
If the pressing force on the slidding surfaces is greater then friction will be
Tema [17]

Answer:

Kinetic friction is lesser than limiting friction. Two surfaces are rubbed together, first with a smaller force and then with a greater force.

3 0
2 years ago
Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo
Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c)                           d)   w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

  •                T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

                T = 1 / f

                 f = 1 / T

                 f = 1 / 3.6 104

                 f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

                 d = 2pi (R + r)

                 d = 2pi (69 911 103 + 120 106)

                 d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

                 .w = 2pi /t

                  w = 2pi / 3.6 10⁴

                  w = 1.745 10⁻⁴ rad / s

how fast is

                  v = w r

                  v = 1.75 10-4 (69.911 106 + 120 106)

                  v = 190017 m / s

5 0
2 years ago
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