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2 years ago

Which statement describes a property of a proton?

Physics

1 answer:

lesantik [10]2 years ago

7 0

Answer:

They have no charge and are present in the nucleus of an atom. They have a negative charge and travel around the nucleus of an atom. They have a positive charge and travel around the nucleus of an atom

Explanation:

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Find B when θ=35, E=10V, t=5, N=250, A=1.20m^2

givi [52]

Answer:

its most definitely c. trust me

Explanation:

3 0

2 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago

HELP ASAP!!!! 20 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

Alexxandr [17]

Answer:

I would have to say the answer is D

Explanation:

because the angle is being changed using the ray box.

8 0

3 years ago

Without fluid friction, all objects accelerate at?

lys-0071 [83]

Answer:

Acceleration of gravity= $9.8m/s/s$

Explanation:

Newton's Second Law-acceleration is proportional to the net force acting on an object.

All objects usually free fall at the same acceleration of $9.8m/s/s$ -this regardless of their mass. This acceleration is known as acceleration of gravity.

7 0

3 years ago

An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts

Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0

2 years ago

Read 2 more answers