What force pulls all objects with mass toward<br> one another?

39 liters of water at

Harman [31]

Answer:

Temperature of the mixture=8 °C

Explanation:

Let, Temperature of the mixture=t °C

Q₁ = Q₂

=> m₁SΔΘ₁ = m₂SΔΘ₂

=> 39 × (20+t) = 21 × (60-t)

∴ t = 8 °C

hope you have understood this

pls mark it as the brainliest

3 0

2 years ago

Needing help with this please

Gemiola [76]

Answer:

So A we cant sadly do because we cant draw. B is going to be kinetic. Thats because static friction means it stays in one place, for kinetic it means moving. So it will be 0.05 as the coefficient of the friction. Sadly, I cannot calculate C. You will have to use trigonemetry but I cannot fit that big an explanation.

Answer to A: the free body diagram would be the ski things inclined with gravity, friction, and air resistance. I except you know which directions

Answer to B: Kinetic friction is the answer.

Answer to C: Find on own, I cannot write super big explanations - use trigonometry.

6 0

2 years ago

A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object

Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

$Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

Em₀ = K = ½ m v²

Em= $Em_{f}$

½ m v² = m g h

v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v

Initial before the crash

p₀ = M v₁₀ + 0

Final after the crash

$p_{f}$ = M v1f + m v

p₀ = $p_{f}$

M v₁₀ = M $v_{1f}$ + m v

As the shock is elastic the kinetic energy is conserved

K₀ = $K_{f}$

½ M v₁₀² = ½ M $v_{1f}$ ² + ½ m v²

Let's write the system of equations

M v₁₀ = M $v_{1f}$ + m v

M v1₁₀² = M $v_{1f}$ ² + m v²

We cleared v1f in the first we replaced in the second

$v_{1f}$ = (M v₁₀ - mv) / M

M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

2 m v₁₀ - v (m + 1) m/ M = 0

v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)

V₁₀ = 7.668 (2.68) / 10.82

v₁₀ = 1.90 m / s

5 0

2 years ago

Which of the following is a common downside of net pens, one of the most common forms of aquaculture?

Eduardwww [97]

Answer:

Answer is overcrowding aka answer choice A. I got the question and got it right. Please mark brainliest. Have a good day! :)

Explanation:

8 0

2 years ago

Read 2 more answers

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$