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zimovet [89]
3 years ago
14

Please help on this one ?

Physics
1 answer:
satela [25.4K]3 years ago
5 0

It i c. because Newton's law was all about how gravity pulls us downward so that was going to be a difficulty in the Apollo mission.

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A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi
Rus_ich [418]
A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² =  1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² =  1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s


b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s² 
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
3 0
3 years ago
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
Write short letters.
Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

Yours affectionately

your name

4 0
2 years ago
Read 2 more answers
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
The main benefit you receive from using locomotor movements in a variety of activities is?
adell [148]

Answer:The locomotor skills include: walking, running, skipping, galloping, hopping, jumping, sliding, walking backwards, and leaping. Students are learning these skills at it could take lots of practice to develop the skills necessary to complete all of the locomotor skills.

Explanation:

7 0
3 years ago
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