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Sedaia [141]
4 years ago
11

A student performs a lab measuring the velocities of toy cars of different masses

Physics
2 answers:
stiv31 [10]4 years ago
5 0
The answer is Car 1 and Car 2.
andreyandreev [35.5K]4 years ago
5 0

Answer:

Car 1 and Car 2

Explanation:

5 2.2 11

5.5 2 11

6 1.35 8.1

6.5 1.9 12.35

7 1.35 9.45

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A foot player runs 1.6m/s and has a KE of 790 J. What is his mass?
Mariana [72]
The equation for kinetic energy is,

Ke = (1/2)mv^2.

You're given a kinetic energy of 790 joules, and a speed of 1.6 m/s. Plugging these values into the equation, we get,

790 = (1/2)(1.6)^2(m).

Solving for m, we get,

m = (790)/(0.5(1.6)^2).

I'll let you crunch out those numbers for yourself :D

If you have any questions, feel free to ask. Hope this helps!
3 0
3 years ago
In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p
Delicious77 [7]
<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
   m1 * v1 + m2 * v2 = m * v
 1 * 1 + 5 * 0 = (1 + 5) * v
  1 = 6 * v
 v = 1/6 m/s
   So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>
6 0
3 years ago
Name nine elements that have been around so long that we don't even know when they
lora16 [44]

Answer:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

Explanation:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

-  (These are the most longest elements that have been around very long )

7 0
3 years ago
Read 2 more answers
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?
gizmo_the_mogwai [7]
It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
5 0
3 years ago
Read 2 more answers
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