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NeX [460]
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?

Physics
1 answer:
egoroff_w [7]2 years ago
5 0

Answer:

YAYA MAN

Explanation:

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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
11 months 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 = πr²v_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
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