Question 2 of 10

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

VikaD [51]

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

$a=\dfrac{F}{m}$

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

5 0

4 years ago

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An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

6 0

3 years ago

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air

Mariulka [41]

Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0

3 years ago

When was hawaii volcanoes national park established

Lerok [7]

1906 i Believe..........

6 0

3 years ago

An electric fan is running on HIGH. After fan has been running for 1.1 minutes, the LOW button is pushed. The fan slows down to

madam [21]

Answer:

ωi = 15.4 rev/sec

Explanation:

Since the movement of the fan is rotating, we are thus dealing with Rotational motion. In rotational motion, for angular speed to take place also means angular acceleration is also occurring.

angular acceleration = α = (change in speed)/(change in time)

angular acceleration = α = Δw/Δt = (ω - ωi) /(t- t₀) ..........(equation 1)

α = (ω -ωi) /(t- 0)

α = (ω-ωi) /t

ωi = ω - αt ......................................(equation 2)

where ωi is the initial angular speed.

We replace the values for ω, t and α

ωi = 105 rad/sec - ( 4.4 rad/sec² )(1.85s) = 96.86 rad/s = 15.415747788 rev/sec

7 0

3 years ago