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Serggg [28]
2 years ago
6

A high school physics student is sitting in a seat read-

Physics
1 answer:
Nataly_w [17]2 years ago
7 0

The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

  • The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

            ∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

            N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}

Let's analyze the forces on the student.

          N_{student} - W_{student} = 0  \\N_{student} = W _{student}

           

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

  • The reaction force that the chair exerts on the student's support is equal to the student's weight.

Learn more here: brainly.com/question/18117041

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11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

approximately 3.33 m/s^2

the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

6 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
3 years ago
A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.
Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

v^{2}=u^{2}+2as

0^{2}=35^{2}- 2 \times 9.8 \times h

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
Q = cmAT
castortr0y [4]

Answer: Q=3000 cal

Explanation:

We are given the following formula:

Q=m. c. \Delta T   (1)

Where:

Q=3000 cal is the amount of heat

m=300g  is the mass  of water

c=1 cal/g \°C  is the specific heat of water

\Delta T  is the variation in temperature, which in this case is  \Delta T=30\°C-20\°C=10\°C  

Rewriting equation (1) with the known values at the right side, we will prove the result is 3000 cal:

Q=(300g)(1 cal/g \°C)(10\°C)   (2)

Q=3000 cal   This is the result

8 0
2 years ago
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