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sashaice [31]
2 years ago
12

A 75 kg man standing in a lift notices that a as the lift rises, the scale reads 825 n. What is the acceleration of the lift?.

Physics
1 answer:
Leviafan [203]2 years ago
7 0

Answer:

mass=75kg

force=825n

we know,

force=acceleration×mass

or, 825=a×75

or,825/75=a

•°•a=11m/s^2

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A vertical wire carries current in the upward direction. an electron is traveling parallel to the wire. what is the angle α betw
Luden [163]
Angle α between the velocity of theelectron and the magnetic field of the wire will be 90 degree
7 0
3 years ago
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration
umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

F=qvB     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

F=qvB=ma       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0
2 years ago
Which of the following statements BEST explains why ball-and-socket joints have the greatest range of motion?
galina1969 [7]

Answer:

its d on edge c:

5 0
2 years ago
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
2 years ago
The time period T of a simple pendulum is given by the relation
Vanyuwa [196]

Answer:

T^2 \propto L

Explanation:

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}

Taking the square of this equation, we get:

T^2 = (2\pi)^2 \frac{L}{g}

So we see that T^2 is proportional to L and inversely proportional to g. So, we can write:

T^2 \propto L\\T^2 \propto \frac{1}{g}

So the only correct option is

T^2 \propto L

5 0
3 years ago
Read 2 more answers
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