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Svetach [21]
1 year ago
8

Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second

Physics
1 answer:
omeli [17]1 year ago
5 0
KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)

KE = (0.5) times the mass times the velocity^2
square the velocity first

Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules
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Answer

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Step-by-step explanation:

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5 0
2 years ago
An electric heater draws a steady current = 20.0 A on a 120-V line. (a) Calculate how much power does it require.
babymother [125]

Answer:

The heater power required is 2400 W. The power in the heater can be calculated as the product of the voltage line and the steady current:

P=V.I

P=120 V * 20 A = 2400 VA = 2400 W

Explanation:

8 0
2 years ago
An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t
Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

\omega = positive

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

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so here correct answer is

B. W is positive and a is negative

8 0
1 year ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P
Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0
2 years ago
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