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Harlamova29_29 [7]
2 years ago
3

The weight of a cart with a mass of 150 kg is ___ ?

Physics
2 answers:
luda_lava [24]2 years ago
6 0

Answer:

Explanation:

Weight is a product of mass and acceleration due to gravity. Weight is also a downward force upon a body.

From newton's second law of motion which states that the rate of change of momentum of a body is proportional to the applied force which takes place in the direction of force applied.

This law gives a formula which relate force, mass and acceleration.

Force = mass x acceleration due to gravity (g)

Given that mass= 150kg , g= 10m/s²

Since force(F) weight = mass(m) x acceleration due to gravity (g)

Therefore W = mg

W= 150kg x 10m/s²

Weight = 1500N

The S. I unit of weight is newton (N)

I hope this was helpful, Please mark as brainliest  

bezimeni [28]2 years ago
5 0

Answer:1500N

Explanation:

Mass=150kg

Acceleration due to gravity =10m/s^2

weight=mass x acceleration due to gravity

Weight =150x10

Weight =1500N

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g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the
Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

R = u^2 / g

5 0
2 years ago
A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal
ASHA 777 [7]
Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
F_{centripetal\:force} =  \frac{m*s^2}{R}

Solving:
F_{centripetal\:force} = \frac{m*s^2}{R}
F_{centripetal\:force} = \frac{68*3.9^2}{6.5}
F_{centripetal\:force} = \frac{68*15.21}{6.5}
F_{centripetal\:force} = \frac{1034.28}{6.5}
\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}
Answer:
<span>B.159 N</span>
3 0
2 years ago
A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
  • According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
  • Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

       = 0.3759

r = Sin^-1 0.3759

   = 22.08

   <u>≈ 22°</u>

3 0
2 years ago
The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end
Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

f_1=150\ Hz

for n=2

f_2=\frac{5v}{4L}

Thus there is one additional resonance corresponding to n=1 , between f_1 and f_2

8 0
2 years ago
your drop a coin from the top of a hundred-story building(1000m). If you ignore air resistance, how fast will it be falling righ
Ede4ka [16]

consider the motion of con from top to bottom

Y = vertical displacement = 1000 m

a = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity at the top = 0 m/s

v = final velocity at the bottom = ?

using the kinematics equation

v² = v²₀ + 2 aY

v² = 0² + 2 (9.8) (1000)

v = 140 m/s


t = time taken to hit the ground

Using the equation

v = v₀ + at

140 = 0 + 9.8 t

t = 14.3 sec

7 0
2 years ago
Read 2 more answers
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