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3 years ago

14

I need emergency help we only have 3 minutes left

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

Option A

Explanation:

Ast the force is equal and the diayance is equal the beam is also balanced

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What is the reaction force of the table with a weight of 558N

Delicious77 [7]

Answer: reaction force = -558N

Explanation:

w = f = 558N

since action force and reaction force are equal in magnitude and opposite in direction,

reaction force = -(f)

reaction force = -558N

if that helps.

8 0

3 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

$D$ = distance between sue and van = 160 m

$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

hence there will be collision

7 0

3 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

SVETLANKA909090 [29]

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$

5 0

3 years ago

You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r

sesenic [268]

Answer:

0.139 rad

Explanation:

We use Snell's law $n_1sin\theta_1=n_2sin\theta_2$ , where if $n_1$ is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, $\theta_1$ would be the <em>incident angle</em>, and if $n_2$ is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, $\theta_2$ would be the <em>refraction angle</em>, which we want, so we do:

$sin\theta_2=\frac{n_1}{n_2}sin\theta_1$

And finally:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)$

We then insert our values:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686
)=0.139 rad$

6 0

3 years ago

Is the maximum displacement of a point on a vibrating body or wave.

Georgia [21]

The correct answer is amplitude

7 0

3 years ago