Ask question

Ask question

All categories

2 years ago

15

an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m

and the mass of the planet is 6.35 x 10^30 Kg estimate g as 6.67 x 10^-11 N (m/kg)

1 answer:

Ksenya-84 [330]2 years ago

3 0

Answer:

7.87x10^5m/s^2

Explanation:

You might be interested in

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

yulyashka [42]

Answer:

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s

3 0

2 years ago

Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0

3 years ago

Is a 10 year old supposed to be doing algebra

forsale [732]

If he feels like, is interested in it, and is able to grasp it, then why not ? Why not indeed ?

3 0

2 years ago

Read 2 more answers

What other kind of simple machine is a special kind of lever?

ehidna [41]

saitama is the strongest

4 0

2 years ago