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Stells [14]
3 years ago
14

The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the

elevator, if upward is the positive direction. Write the numerical value only with 2 decimal places. Do not write the unit. If the acceleration is negative (downward) include the negative sign in your answer.
Physics
1 answer:
Ronch [10]3 years ago
8 0

Answer: 1

Explanation:

Given

Tension is the string T=44\ N

mass of object m=4\ kg

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2

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A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10
anzhelika [568]

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by  Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90°  = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.

5 0
3 years ago
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
4 years ago
A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov
lions [1.4K]

Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ =  3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

                           = 9 - 14.8 + 14

                           = 8 .2 kg m/s

hence, the total momentum is 8 .2 kg m/s in north direction.

7 0
4 years ago
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley
FinnZ [79.3K]

awnser :

Explanation:

8 0
3 years ago
Read 2 more answers
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