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Sedaia [141]
3 years ago
12

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the

mass of the object is
Physics
2 answers:
pantera1 [17]3 years ago
8 0
Newton's second law states that Fnet = ma, where Fnet is the net force applied, m is the mass of the object, and a is the object's acceleration. You have the values for Fnet and a, so you simply use this equation to solve for m, mass.
Darina [25.2K]3 years ago
4 0

Answer:

m=5.2kg

Explanation:

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the mass of the object is

from newton's second law of motion , which states that

the rate change in momentum of an object is directly proportional to the force applied

definition of terms  you should know.

a force is tat which tends to cane a body state of rest or uniform motion in a strait line

mass is the quantity of matter in a body

acceleration is change in velocity per time

therefore,

f∝m(V-U)/t

f=km(V-U)/t

k=1

f=m(V-U)/t

(V-U)/t=a

f=ma

f=125N

a=24m/s^2

m=5.2kg

the mass of the object

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An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel
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Answer:

L = 0.635m

Explanation:

This problem involves the concept of stationary waves in pipes. For pipes closed at one end,

The frequency f = nv/4L for n = 1,3,5....n

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f = nv/2L for n = 1,2,3,4...n

Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.

The film solution can be found in the attachment below.

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3 years ago
9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be
Akimi4 [234]

Answer:

525 Bq

Explanation:

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A = A₀ (½)^(t / T)

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A₀ is the initial amount,

t is the time,

T is the half life

A = (8400 Bq) (½)^(18.0 min / 4.50 min)

A = (8400 Bq) (½)^4

A = (8400 Bq) (1/16)

A = 525 Bq

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The Kyoto protocol works by _______.
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Read 2 more answers
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
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