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Lorico [155]
3 years ago
8

Two difference between thrust and upthrust .

Physics
1 answer:
velikii [3]3 years ago
6 0

Upthrust is the upward force exerted by fluids on the surface of an object immersed in fluids. Thrust:- It is the force acting perpendicular to the surface. Upthrust:- It is the upward force exerted by the fluid on the surface of an object immersed in liquid.

I hope it's help you

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15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m
Rudik [331]

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters



7 0
3 years ago
Read 2 more answers
In SI units, what is the magnitude the net force acting on a 1,152 kg car that accelerates uniformly along a straight line from
geniusboy [140]

Answer:

Fnet = 14515.2 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force.
  • Fapp is the applied force.
  • Fg is the force due to gravitation.

Given the following data;

Mass = 1,152 kg

Initial velocity, u = 3m/s

Final velocity, v = 17m/s

Time, t = 5 seconds

To find the magnitude of the net force;

First of all, we would determine the acceleration of the car.

Acceleration = (v-u)/t

Acceleration = (17 - 3)/5

Acceleration = 14/5

Acceleration = 2.8m/s

To find the applied force;

Fapp = mass * acceleration

Fapp = 1,152 * 2.8

Fapp = 3225.6 N

Next, we would find the force exerted on the car due to gravity.

Fg = mass * acceleration due to gravity

We know that acceleration due to gravity is equal to 9.8m/s²

Fg = 1152 * 9.8

Fg = 11289.6N

Substituting the values into the equation, we have;

Fnet = 3225.6 + 11289.6

Fnet = 14515.2 Newton

7 0
3 years ago
An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista
ioda

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

6 0
3 years ago
Read 2 more answers
An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta
icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

v=\sqrt{\frac{T}{\mu} }

where T= 200 N is tension in the string , \mu=1.0 g/m is the linear mass density

v=\sqrt{\frac{200}{1\times 10^{-3} }

v=447.2 m/s

Wavelength of the wave in the string is

\lambda =2L=2\times 0.8=1.6 m

The frequency is

f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m

8 0
3 years ago
Which one of the following statements about attitudes is correct?
ozzi
B is a correct answer. Positive attitudes can help us learn quicker. Hope it helped you, and have a great day.

-Charlie
7 0
3 years ago
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