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3 years ago

The diagram shows a rectangular object acted on by three horizontal forces.

Physics

2 answers:

GalinKa [24]3 years ago

5 0

I think the answer is B

sammy [17]3 years ago

4 0

<h3>right --- 20 + 10 = 30 N</h3>

<h3>resultant force --- 30 - 10 = 20 N</h3>

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Which force can affect an object without direct physical contact?

denis-greek [22]

Gravity affet everything and it touches nothing.
Hope this helps!

8 0

4 years ago

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A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center

uranmaximum [27]

Answer:

4.36 rad/s

Explanation:

Radius of platform r = 2.97 m

rotational inertia I = 358 kg·m^2

Initial angular speed w = 1.96 rad/s

Mass of student m = 69.5 kg

Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2

Therefore initial rotational momentum of system = w( Ip + Is)

= 1.96 x (358 + 613.05)

= 1903.258 kg.rad.m^2/s

When she walks to a radius of 1.06 m

I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2

Rotational momentuem of system = w(358 + 78.09) = 436.09w

Due to conservation of momentum, we equate both momenta

436.09w = 1903.258

w = 4.36 rad/s

7 0

3 years ago

An object is acted upon by the forces F1equalsleft angle12,7,2right angle and F2equalsleft angle0,4,8right angle. Find the f

Umnica [9.8K]

Answer:

$F_3=(-12,-11,-10)$

Explanation:

It is given that,

Force, $F_1=(12,7,2)$

Force, $F_2=(0,4,8)$

We need to find the force F₃ that must act on the object so that the sum of the forces is zero. Let $F_3=(x,y,z)$

According to question, $F_1+F_2+F_3=0$

$(12,7,2)+(0,4,8)+(x,y,z)=0$

$(12+x)+(11+y)+(10+z)=0$

Since, 12 + x = 0, 11 + y = 0 and 10 + z = 0

The third force is equal to,

$F_3=(-12,-11,-10)$

Hence, this is the required solution.

4 0

3 years ago

What is the wavelength of a radiowave that has a frequency of 9.650 x10^7 Hz (cycles per second)?h = 6.6261 x10^-34 J/sc = 2.998

slava [35]

Hello!!

For this, we only need frequency and speed, so for calculate we applicate formula:

$\boxed{\lambda = V/f}$

$\textbf{Being:}$

$\sqrt{}$ $\lambda = Wavelength = \ ?$

$\sqrt{}$ $V = Velocity = 2.998*10^{8} \ m/s$

$\sqrt{}$ $f = Frequency = 9,650*10^{7} \ Hz$

$\text{Then let's \textbf{replace it according} we information:}$

$\lambda = 2,998*10^{8} \ m /s / 9,650*10^{7} \ Hz$

$\lambda = 3.106 \ m$

$\textbf{Result:}\\\text{The wavelength of the radiowave is \textbf{3.106 meters}}$

4 0

3 years ago

Consider the case in which the object is between the lens and the focal point. Trace the P ray (use the label P1 for the segment

Pani-rosa [81]

Answer:

See attachment

Explanation:

The diagram attached is a sketch of rays when the object is placed between the lens and focal point. As can be seen, the image in such cases is always on the same side of lens as the object. It is formed beyond the focal point from the lens and is magnified

8 0

3 years ago