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taurus [48]
3 years ago
11

Sand is poured at a constant rate of 75.5 g/s into a 0.540-kg bucket resting on a scale. If sand is landing in the bucket with a

speed of 3.70 m/s, determine the scale reading when there is 0.550 kg of sand in the bucket.
Physics
1 answer:
worty [1.4K]3 years ago
4 0

Answer:10.961 N

Explanation:

Given

mass flow rate in bucket is \dot{m}=75.5 g/s

mass of bucket m=0.540 kg

velocity of sand v=3.70 m/s

Force acting on bucket due to incoming of sand is given by

F=\frac{\mathrm{d} m}{\mathrm{d} t}v

F=75.5\times 10^{-3}\times 3.7=279.35\times 10^{-3} N

Now weight of sand and bucket when 0.55 kg sand is Present in bucket

W_1=(0.54+0.55)\times 9.8

W_1=1.09\times 9.8=10.68 N

Total reading on scale will be addition of weight and force acting by incoming sand

Reading =10.68+0.279=10.961 N

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Reil [10]

Answer:

23.67 m

Explanation:

We are given;

Frequency; f = 0.3 Hz

Speed; v = 7.1 m/s

Now, formula to get the wavelength is from the wave equation which is;

v = fλ

Where λ is wavelength

Making λ the subject, we have;

λ = v/f

λ = 7.1/0.3

λ = 23.67 m

5 0
3 years ago
How do I solve the equation
ioda
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6 0
3 years ago
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
Verdich [7]

The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

F=2.4\cdot 10^2 N

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
P wave for 100km time= distance/speed , T=100km/6.1km/s
shutvik [7]
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8 0
3 years ago
An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what
Brums [2.3K]

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

(Distance)S=600m

(acceleration)a= ?

now,

using formula,

45^2=0^2+2×a×600

2025= 1200a

a=2025÷1200

a = 1.68m/s^2(The unit of acceleration is m/s^2)

Therefore the acceleration is 1.68m/s^2

Hope it works !!!

3 0
3 years ago
Read 2 more answers
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