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klasskru [66]
3 years ago
10

A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire

ction of the force on the charge
Physics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

The magnitude of the force is 12 N Upwards

Explanation:

The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.

Given;

magnitude of charge, q = 0.06 C

magnitude of electric field, E = 200 N/C

The magnitude of the force is given by;

F = qE

F = 0.06 x 200 N/C

F = 12 N Upwards

Therefore, the magnitude of the force is 12 N Upwards

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A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
denpristay [2]
Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u


Mb=Ma
48=6u
u=8m/s
3 0
3 years ago
Sound travels through air at 343 m/s,
sasho [114]

The sound wave will have traveled 2565 m  farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m  farther in water than in air.

5 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas
Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

V_{max} =\omega*A=  \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s

(c) the maximum acceleration of the object

a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2

6 0
3 years ago
List five examples from daily life in which you see periodic motion caused by a spring or cord.
dmitriy555 [2]

car suspension

guitar string

clips for hanging clothes

keyboards buttons

elevators

4 0
3 years ago
Read 2 more answers
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