All categories

3 years ago

The main equation we are using to measure the e/m ratio is:

Physics

1 answer:

kow [346]3 years ago

6 0

Answer:

Explanation:

the electrons beams deflection radius will be halved.

You might be interested in

Which statement correctly describes gravity?

Advocard [28]

Answer:

a force that attracts matter to the earth

Explanation:

depends on where you are the gravity can be different in space there is no gravity on Earth there is , that's why when you jump you come back down

5 0

3 years ago

A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m$

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

$\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds$

The reaction time cannot be more than 1.69515 seconds

4 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm?

Stolb23 [73]

-0.012

Done !!!!!!!!

4 0

3 years ago

Car a with mass 1,783 kg collides with stationary 1600 kg car b. they become locked together after the collision and move with s

ValentinkaMS [17]

The initial speed of car A is 15.18 m/s.

Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.

p=p'

pa+pb = pa'+pb'

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

ma = mass of object A (kg) = 1,783 kg
mb = mass of object B (kg) = 1,600 kg
va = speed of object A before collides (m/s)
va' = speed of object A after collides (m/s) = 8 m/s
vb = speed of object B before collides (m/s) = 0 m/s
vb' = speed of object B after collides (m/s) = 8 m/s
p = momentum before collision (Ns)
p' = momentum after collision (Ns)

(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)

(1,783×va) + 0 = 14,264+12,800

(1,783×va) = 27,064

$va \:=\: \frac{27,064}{1.783}$

va = 15.18 m/s

Learn more about The law of momentum conservation here: brainly.com/question/7538238

#SPJ4

3 0

1 year ago