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Fittoniya [83]
3 years ago
6

A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1

.35s. We can ignore air resistance.
Kinematics formula would be most helpful to solve for the target unknown?

Answer choices in picture.

Physics
1 answer:
White raven [17]3 years ago
7 0

Answer:

v_f = v_i + at

v_f = 13.23 m/s

Explanation:

Height Of the watermelon when it is dropped is given as

h = 80 m

time of fall under gravity

t = 1.35 s

now if water melon start from rest then we have

v_i = 0

acceleration due to gravity for watermelon

a = 9.81 m/s^2

now we need to find the final speed of watermelon

v_f = v_i + at

so we will have

v_f = 0 + (9.81)(1.35)

v_f = 13.23 m/s

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Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m
trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ =  30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction  x-axis :

∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T:  cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends  vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends  vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0
3 years ago
Jupiter is 5.2 AU from the sun on average. How long is one year on Jupiter?
velikii [3]

Answer:

12 years

Explanation:

12 years is correct because how long is Jupiter one year is 12 years

3 0
2 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
2 years ago
Can some one help me ;-;
enyata [817]

Answer: The first answer for the first problem, and the 2nd answer for the second problem

Explanation: For the first one, if it is absolute zero, the molecules would not move at all.

For the second one, the temperature of the sample will increase due to the movement.

6 0
3 years ago
An integrated circuit is a ___.
EastWind [94]
HI buddy!

An integrated circuit(IC) is a semiconductor water on which thousands of tiny resistors, transistors are fabricated.

Remember that an integrated circuit are the heart and brains of most circuits. They are the keystone of modern electronics.

I hope this helps!
5 0
2 years ago
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