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What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

$F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N$

The net power needed to change the speed of the vehicle is calculated as;

$P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W$

3 0

2 years ago

If this atom has a balanced charged, how many protons would you expect to find in this atom?

avanturin [10]

The answer would be C because there is six electrons and so there will be six protons because the amount of protons and electrons have to be the same otherwise it would be an unbalanced particle and you wouldn't be able to touch the object it is in without worrying about something happening

3 0

2 years ago

Read 2 more answers

The graph represents the change in that occurs when food is cooked over a charcoal grill. Which statement correctly explains the

MArishka [77]

Option C is the answer

<h3 /><h3>Small Explanation:-</h3>

The reactants are charcoal that is unlit + oxygen and the products are the burnt charcoal + energy.

(Explanation with formula and reason attached. Check it.)

$> Benjemin360$

6 0

2 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

2 years ago

What is the minimum mass needed for a star to be on the main sequence? What happens to stars that do not meet the minimum mass?

Mariana [72]

Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.

To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.

6 0

2 years ago

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