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3 years ago

PLZ HURRY

Physics

1 answer:

anygoal [31]3 years ago

6 0

Answer:

It should just be A

Explanation:

I dont see the difference between these 2 but ill choose A and update you

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Please help with this!!!!!

34kurt

(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped

4 0

3 years ago

Light propagating in the glass n1= 1.65 wall of an aquarium tank strikes the interior edge of the wall with incidence angle 19.0

jonny [76]

We can solve this using Snell's Law which is represented by the equation:
sin θ₁ / sin θ₂ = n₂ / n₁

From the problem, we can substitute values and solve for the angle of refraction.
sin 19 / sin θ₂ = 1.65 / 1
θ₂ = 11.38°

The angle of refraction would be 11.38°.

3 0

3 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

3 years ago

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

valina [46]

The speed of water can be split into vertical and horizontal speed components:
$v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta$

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
$y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t$
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
$-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3$

8 0

3 years ago

A variable that is not changed.

AleksAgata [21]

Answer:

A controlled variable does not change during a experiment

Explanation:

it's c

5 0

3 years ago