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Natalka [10]
2 years ago
5

Four model rockets are launched in a field. The mass of each rocket and the net force acting on it when it launches are given in

the table below. (1. 4.25kg 120N) (2. 3.25kg 120N) (3. 5.50kg 120N) (4. 4.50kg 120N), Which rocket has the highest acceleration?
Physics
1 answer:
kondor19780726 [428]2 years ago
7 0

Answer:

Rocket 2 has highest acceleration.

Explanation:

Net force,

F = mass (m) × acceleration (a)

We have,

m₁ = 4.25 kg and F = 120 N

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{120}{4.25}\\\\a_1=28.23\ m/s^2

m₂ = 3.25 kg, F₂ = 120 N

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{120}{3.25}\\\\a_2=36.92\ m/s^2

m₃ = 5.5 kg, F₃ = 120 N

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{120}{5.5}\\\\a_3=21.81\ m/s^2

m₄ = 4.5 kg, F₄ = 120 N

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{120}{4.5}\\\\a_4=26.66\ m/s^2

Hence, it can be seen that the highest acceleration is of rocket 2.

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a child stands with each foot on a different scale. The left scale reads 200N and the right scale reads 150N. what is her mass i
vitfil [10]

Answer:

35.7 kg

Explanation:

We are given that

Left scale reads =200N

Right scale reads=150 N

We have to find the mass in kg.

We know that

Weight  of child= Sum of two scales=200+150=350 N

Acceleration due to gravity=9.8 m/s^2

We know that

Weight= mg

Using the formula

350=9.8 m

m=\frac{350}{9.8}=35.7 kg

Hence, the mass of child=35.7 kg

4 0
3 years ago
What has a higher eccentricity - a planet or a comet?
Daniel [21]

Answer:

comet

Explanation:

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2 years ago
Titan Tommy and the Test Tubes at a nightclub this weekend. The lead instrumentalist uses a test tube (closed-end air column) wi
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6 0
3 years ago
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
3 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
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