Ask question

Ask question

All categories

2 years ago

12

How much power is developed in a car that can pull a 5000 N at a constant speed of 20 m/s?

1 answer:

egoroff_w [7]2 years ago

5 0

Answer:

YAYA MAN

Explanation:

You might be interested in

lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu

Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

$W=Fd\cos\theta$

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

$\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}$

Therefore, the total work done is 578.123 J and the answer is option E

6 0

1 year ago

Grasses, shrubs and trees are called producers because they make ______a. water_____b. carbon dioxide_____c. minerals_____d. foo

LekaFEV [45]

B?
Carbon Dioxid? Or is is c?

4 0

3 years ago

Read 2 more answers

1. Johnny wants to know where the water line is in a dark well. He drops a penny into the well and counts until he hears the pen

777dan777 [17]

Answer:

3 feet down

Explanation:

i think

8 0

3 years ago

ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,

Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

$d = \frac{(v_f + v_i)}{2} \times t$

here we know that

$v_f = 10 m/s$

$v_i = 0$

$t = 1.2 s$

now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

4 0

2 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago