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aalyn [17]
1 year ago
11

A 0.56 kg model rocket produces 53 newtons of upward thrust as it shoots upward off the launch pad. With what acceleration meter

s per second-squared does the rocket launch?
Physics
1 answer:
Yanka [14]1 year ago
6 0

resultant force = thrust – weight

acceleration = resultant force (newtons, N) divided by mass (kilograms, kg).

Acceleration = resultant force divided by mass

53N/0.56

=94.64 approximately 95

= 95m/s^2

This means that, every second, the speed of the rocket increases by 95m/s2

the S.I unit of Acceleration is meter per second square.

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A brick of mass 2 kg is dropped from a rest position 5 m above the ground. what is its velocity at a height of 3 m above the gro
Rina8888 [55]
We can solve the problem by using the law of conservation of energy.

Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
E= U = mgh=(2 kg)((9.81 m/s^2)(5 m)=98.1 J

when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
E= K+U= \frac{1}{2} mv^2 + mgh'

where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
v= \sqrt{ \frac{2(E-mgh')}{m} }=6.3 m/s
8 0
2 years ago
A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a
JulijaS [17]

All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

8 0
2 years ago
Read 2 more answers
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is
finlep [7]

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

4 0
2 years ago
Read 2 more answers
An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of
Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 /  .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B)  Rest length of the rocket = length of platform = 90 m

C ) length of platform  as viewed by moving observer =

\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }

= \frac{90}{\sqrt{1-\frac{0.92}{1 } } }

= 321 m

D )  For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }

8.75 x 10⁻⁸ s .

8 0
2 years ago
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