A book moving across a desk has a net force equaling 60 N, to the right.

The train traveled 500 kilometers north to Odessa in 2 hours . What’s the train’s speed ? What is the velocity ?

ElenaW [278]

Explanation:

Speed is distance over time.

500 km / 2 hr = 250 km/hr

Velocity is speed and direction.

250 km/hr north

7 0

3 years ago

An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh

melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.

7 0

3 years ago

Which of the following is Not true about the noble gases

Alexandra [31]

Noble gases are not highly reactive

3 0

3 years ago

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c

Romashka [77]

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

6 0

3 years ago

A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

belka [17]

Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

Explanation:

It is given that,

Charge, $q=1.5\ \mu C=1.5\times 10^{-6}\ C$

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

$U=\dfrac{1}{2}q\times V$

$U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36$

U = 0.000027 J

$U=2.7\times 10^{-5}\ J$

So, the potential energy is stored in the capacitor is $U=2.7\times 10^{-5}\ J$ . Hence, this is the required solution.

4 0

3 years ago

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