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Mnenie [13.5K]
2 years ago
15

2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so

Physics
1 answer:
evablogger [386]2 years ago
5 0
The answer is (a) hope it helped!<3






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While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b
Citrus2011 [14]

Answer:

\theta = 66.7 degree

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

F_n = mg + Fsin\theta

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

Fcos\theta = \mu F_n

F cos\theta = \mu (mg + F sin\theta)

F(cos\theta - \mu sin\theta) = \mu mg

F = \frac{\mu mg}{cos\theta - \mu sin\theta}

so here we can say

cos\theta - \mu sin\theta > 0

tan\theta = \frac{1}{\mu}

\theta = tan^{-1}\frac{1}{\mu}

\theta = tan^{-1}(\frac{1}{0.43})

\theta = 66.7 degree

6 0
3 years ago
Does an infrared wave or an x-ray travel faster in the vacuum of space?
Aneli [31]
Ok no se si te puedo
3 0
2 years ago
Read 2 more answers
Please help me out id really appreciate it
Aliun [14]

Answer:

Stroke is the correct answer

8 0
2 years ago
A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th
Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

6 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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