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Lapatulllka [165]
3 years ago
10

Someone help me with this ASAP!!!!

Physics
2 answers:
wlad13 [49]3 years ago
6 0
Yeah the last one seems accurate to me.
laila [671]3 years ago
5 0
I’m thinking the Last one tbh
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Citrus2011 [14]

Instead of asking this question go and study

5 0
2 years ago
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Which of these describes a real image?
Margaret [11]
Image from a far away object formed by a concave mirror

I have no idea but this is my best guess as a sophomore in college
8 0
3 years ago
Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0
3 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0
3 years ago
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