WHAT DOES THE EXPRESSION "EM" IN EM SPECTRUM REPRESENT

In Figure 4.24, a current of 0.3 A flows through the conductor CD, and a charge of 4C passes through a cross-section AB of the

olga2289 [7]

Answer:

13.33 seconds

Explanation:

I = Q/t

t = Q/I = 4/0.3 = 13.33 seconds

8 0

3 years ago

Read 2 more answers

What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.

liq [111]

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

$KE=\frac{1}{2}mv^2$

where m is the mass in kilograms and v is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

m=50

v=60

Substitute these values into the formula.

$KE=\frac{1}{2}*50*60^2$

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

$KE=\frac{1}{2}*50*3,600$

Multiply 50 and 3,600

$KE=\frac{1}{2}*180,000$

Multiply 1/2 and 3,600, or divide 3,600 by 2.

$KE=90,000$

Add appropriate units. Kinetic energy uses Joules, or J.

$KE=90,000 Joules$

The kinetic energy of the object is 90,000 Joules

6 0

3 years ago

If you walk at an average speed of 5 km/h for 30 minutes, how

Inessa05 [86]

The distance that would be accumulated during the journey is 2.5 meters

The parameters given in the question are written below;

average speed= 5 km/hr

time = 30 minutes

convert 30 minutes to hours

= 30/60

= 0.5 hours

Distance-= speed × time

= 5 × 0.5

= 2.5 meters

Hence the distance of the entire journey is 2.5 meters

Please see the link below for more information

brainly.com/question/24268730?referrer=searchResults

3 0

2 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago