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crimeas [40]
3 years ago
9

What is one characteristic of an electron?

Physics
2 answers:
Varvara68 [4.7K]3 years ago
8 0

Answer:

the answer to this question is A

inysia [295]3 years ago
3 0

Answer:

D. Almost no mass

Explanation:

apex

You might be interested in
A concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35°. The kinetic between the slab and slope is 0.60
belka [17]

I got you

Explanation:

normal force = 400 g cos 35

friction force up slope = .6 (400 g) cos 35

weight component down slope = 400 g sin 35

400 a = 400 g sin 35 - .6 (400 g cos 35)

a = g (sin 35 - .6 cos 35) = .082 g

I hope this helps you

8 0
2 years ago
A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s
myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder =  150 mm

thickness of cylinder = 6 mm

maximum shear stress =  25 MPa

we know that

hoop stress is given as =\frac{pd}{2t}

axial stress is given as =\frac{pd}{4t}

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}

25 = \frac{pd}{8t}

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0
3 years ago
(8c5p80) Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward f
Alexus [3.1K]

Answer:

The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.

Explanation:

The engine of the craft provides an upward thrust of 3480 N so that the space craft descends at a constant speed.

This implies that the net force on the space craft is zero.

The upward thrust will be equal to the downward gravitational pull by Callisto.

So the weight of the craft near the vicinity will be 3480 N.

3 0
3 years ago
Question 3 Please, i need help! Thank you
Tom [10]
I don't know this 1 I'm sorry I can't help you 
4 0
3 years ago
QUESTION 7
ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+at

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

\displaystyle a=\frac{0-20.4}{7.4}

a=-2.757\ m/s^2

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F = 1,126\ kg * 2.757\ m/s^2

F= 3,104 N

The force required is 3,104 N

6 0
3 years ago
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