Pls help if u do it 100 points

On the earth, when an astronaut throws a 0.250 kg stone vertically upward, it returns to his hand a time T later. On planet X he

Igoryamba

Answer: $\frac{g}{2}$

Explanation:

Given

mass of stone $\left ( m\right )$ =0.250 kg

Let initial velocity with which it is thrown upward is u

therefore after time t it's velocity is zero at highest point

t= $\frac{u}{g}$

where g= gravity at earth

therefore $T=2\times \frac{u}{g}$ -------1

Now same thing is done in Planet X where gravity is g'

therefore time taken by stone to reach surface is

$T'=T=2\times \frac{u}{g'}$ -------2

Divide 1 & 2

$\frac{T}{T'}$ = $\frac{g'}{g}$

$\frac{1}{2}$ = $\frac{g'}{g}$

g'= $\frac{g}{2}$

8 0

2 years ago

A student applies a 10 N force to a wood block with a mass of 5 kg. The block is pushed across four different surfaces. The acce

Furkat [3]

The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?

Answer: The acceleration of the block is $2 m/s^{2}$ .

Explanation:

Given: Force = 10 N

Mass = 5 kg

It is known that force applied on an object is the product of mass and acceleration.

Mathematically, $F = m \times a$

where,

F = force

m = mass

a = acceleration

Substitute the values into above formula as follows.

$F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}$

Thus, we can conclude that the acceleration of block is $2 m/s^{2}$ .

8 0

2 years ago

A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

Aloiza [94]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

$f=\dfrac{R}{2}$

$f=\dfrac{-31}{2}$

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

4 0

2 years ago

For the following inclined plane, m1 = 5.0 kg, m2 = 6.0 kg, and a = 30° A. Draw force diagrams for blocks m1 and m2. B. Calculat

sattari [20]

Answer:

See attached

Explanation:

6 0

2 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

2 years ago