1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
liraira [26]
4 years ago
8

In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak

e observations and collect data regarding the motion of these objects. In the space below, write a general scientific question that you will answer by doing this experiment.
Physics
2 answers:
polet [3.4K]4 years ago
7 0

How can we experimentally verify newton's laws?

borishaifa [10]4 years ago
3 0

Answer:

How can Newton's laws be experimentally verified?

Explanation:

Just took it right now

You might be interested in
An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N, find the coefficient of kinetic
Stolb23 [73]

Answer:

The coefficient of  kinetic friction = 0.026  

Explanation:

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.

Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.

The external force acting on the body is in the forward direction and the friction acts in the backward direction.

Friction increases as the mass of the body increases.

Friction = u_{k}\times m \times g

We now equate this to the external force of 15 N.

15 = u_{k} \times 56 \times 10

u_{k} = \frac{15}{560}

u_{k} = 0.026

The coefficient of  kinetic friction = 0.026  

8 0
4 years ago
Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00
deff fn [24]

Answer:

c < a = b

Explanation:

The tensile stress = Force applied/(Cross sectional area)

(a) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(b) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(c) The applied force = 100 N

The cross sectional area = 2.00 mm² = 2 × 10⁻⁶ m²

The tensile stress = 100 N/(2 × 10⁻⁶ m²) = 50,000,000 Pa = 50 MPa

Therefore, the tensile stress from smallest to largest are;

(a) 50 MPa, < (b) 200 MPa = (a) 200 MPa

Therefore, we have;

c < a = b.

8 0
3 years ago
Please help on this one
Murljashka [212]
Just find the area of the graft

4m/s x 5s
=20m

7 0
3 years ago
Read 2 more answers
7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How
Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

Y = \frac{FL}{Ae}

<u></u>

<u>Part A</u>

cross sectional area and applied force are the same as the original but the length is doubled

\frac{FL_1}{A_1e_1} =  \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} =  \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o

The second wire will be stretched by 2 mm

<u>Part B</u>

a third wire with the same length but twice the diameter of the first

\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

3 0
4 years ago
If Charge A is positive, then Charge B must be:
Eduardwww [97]
It must be positive also
Explanation The energy is coming out of A back into b and then going into A again and out
4 0
3 years ago
Other questions:
  • A zoologist standing on a cliff MJ tranquilizer gun at a monkey hanging from a distance tree branch. The barrel of the gun is ho
    5·1 answer
  • What organisms would have the best likelihood of surviving the bombardment of
    6·2 answers
  • When observing an object, you determine it is moving at a constant velocity and in a straight line. Which of these BEST describe
    9·2 answers
  • When the volume is held constant how does the pressure change as temperature increases? PLEASE HELP
    9·1 answer
  • Two mechanical waves that have positive displacements from the equilibrium position meet and coincide. What kind of
    13·1 answer
  • How many vibrations per second are associated with a 101-mhz radio wave?
    15·1 answer
  • Imagine two 2.0 g bags of protons, one at the earth's north pole and the other at the south pole. part a how many protons are in
    9·1 answer
  • The frequency of the sound heard by the motorcyclist is given by the expression
    11·1 answer
  • The is greater for low-mass stars than it is for high-mass stars. 2. The stars known as are the very largest and brightest of al
    7·1 answer
  • Question 2 of 10
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!