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SpyIntel [72]
3 years ago
13

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 85 N when the elevator is

standing still.
a) What is the reading when the elevator is moving upward with a constant speed of 7.9 m/s?

b) What is the reading when the elevator is moving upward with a speed of 7.9 m/s while decelerating at a rate of 1.1 m/s2?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

Explanation:

Balance force ( W ) = 85 N

a)  moving upward with constant velocity (V ) = 7.9 m/s.

The reading will not change since it is moving with constant velocity

  W = 85 N

b) Moving upward with V = 7.9 m/s and decelerating at the rate of (a ) = 1.1 m /s²

The net force ( F ) = W - ma

where m = W/g = 85/9.81 = 8.6646 kg

F = 85 - 8.6646 * 1.1 = 75.4689 N

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Maslowich

Answer:

.a = 849.05 m / s²

Explanation

The centripetal acceleration is

            a = v² / r

     

Linear and angular velocity are related

          v = w r

Angular velocity and frequency are related by

        w = 2π f

Let's replace

        a = w² r

         a = 4π² f² r

Let's reduce to the SI system

       f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

       .r = 10.3 cm = 0.103 m

Let's calculate

       a = 4π² 14.45²  0.103

       .a = 849.05 m / s²

8 0
4 years ago
What is the path that an electric current follows called
Gwar [14]
I’m sure it’s called a circuit:)
5 0
3 years ago
Read 2 more answers
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
1 point
alexandr1967 [171]

Answer:

length

<em>Your</em><em> </em><em>well</em><em> </em><em>wisher</em><em> </em><em>:-)</em>

4 0
3 years ago
Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an
Paladinen [302]

Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F=m \times a

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

  • m= 8 kg
  • a= 1.6 m/s²

Substitute these values into the formula.

F= 8 \ kg * 1.6 \ m/s^2

Multiply.

F= 12.8 \ kg*m/s^2

<h3>2. Acceleration of the 2nd Object </h3>

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

F= m \times a

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

  • F= 12.8 kg *m/s²
  • m= 2 kg

Substitute the values into the formula.

12.8 \ kg*m/s^2= 2 \ kg *a

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}

\frac {12.8 \ kg*m/s^2}{2 \ kg}=a

The units of kilograms cancel.

\frac {12.8}{2}\ m/s^2=a

6.4 \ m/s^2=a

The acceleration is 6.4 meters per square second.

4 0
3 years ago
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