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ycow [4]
3 years ago
9

A hovercraft is being driven across a lake on a very windy day. The wind exerts a force of 5000 N north on the hovercraft. The p

ropellers driving the hovercraft also exert a force.
What is the net force (to 3.s.f ) on the hovercraft when the force applied to the hovercraft by the propellers is: 10 000 N west
Physics
1 answer:
krok68 [10]3 years ago
4 0

Answer:

The net force on the hovercraft is 11200 N.

Explanation:

Given;

force exerted on the hovercraft by wind, F₁ = 5000 N north

force exerted on the hovercraft by the propeller, F₂ = 10,000 N west

The net force on the hovercraft is calculated as;

F_{net} = \sqrt{F_1^2 + F_2^2} \\\\F_{net} = \sqrt{5000^2 + 10,000^2} \\\\F_{net} = 11180.34 \ N\\\\F_{net} = 11200 \ N \ (3.s.f)

Therefore, the net force on the hovercraft is 11200 N.

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A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
ser-zykov [4K]

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

v = 45m/s

t = 5s

Therefore,

d = v*t

= 45*5

= 225m

5 0
3 years ago
A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the
Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

4 0
3 years ago
A group of cells working together make up a(n)
Neko [114]
Answer: Tissue.

Molecules make up cells, cells make up tissue, tissue makes up organs, organs make up organ systems. 
6 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

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s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
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What the kinetic energy quantities in calculation pls help me​
Rashid [163]

Answer:

KE = 0.5 * m * v², where: m - mass, v - velocity.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2.

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3 years ago
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