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1 year ago

5

If the same force is applied to an object with a small mass it will have a

1 answer:

Rashid [163]1 year ago

8 0

Answer:

acceleration

Explanation:

i hope this helps

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Both formal and informal

Tema [17]

Explanation:

is this a question????...

5 0

2 years ago

Which wave has a longer period and how many times, the wave with a frequency of 7000Hz or the wave with a frequency of 21.000Hz?

Nuetrik [128]

Answer:

I would say the answer is the wave of 21.000Hz

Explanation:

Because it has more frequency, and as more frequency you add, the time or longer period also increases.

3 0

2 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$

To find Ny, we need to find the tension T.
For this, we can equate the net horizontal force.

$F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N$

Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

$N_y= (40*9.8)-(169.8*sin59)=246.4N$

Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

$N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N$

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0

1 year ago

Read 2 more answers

Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550

Arisa [49]

Answer:

q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2 (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

w_poly = R*(T_f - T_i)/(1-n)

w_poly = 0.189*(350 - 400)/(1-1.2)

w_poly = 47.25 KJ/kg

-Hence,

q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

q_poly = 14.55 KJ/kg

4 0

2 years ago

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m

umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

6 0

2 years ago

Read 2 more answers