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Naily [24]
2 years ago
7

Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the h

igher track? _________
a. cars travel faster on the lower track
b. cars travel faster on the higher track
c. There is no difference in speed

How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is__________
a. Shorter
b. Longer
c. Equal

Calculate speeds for each track. How much faster was the car on the higher track than the lower track?__________
a. .56 m/s
b. 1.54 m/s
c. 3.50 m/s

Physics
2 answers:
raketka [301]2 years ago
7 0

Answer:

1.a

2.longer

Explanation:

nata0808 [166]2 years ago
6 0

Answer:

1.Cars Travel faster on higher track

2.Shorter

3.0.56 m/s

Explanation:

I just took the assignment on e2020

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A 1 mg ball carrying a charge of 2 x 10-8 C hangs from a
Fed [463]

Answer:

σ = 0.255*10^-3 C/m²

Explanation:

The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.

At equilibrium ,

Upward force = downward force

Tcosθ = mg ----(1)

Assuming that the Forward force = backward force, then

Tsinθ = σq/ε₀

[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)

Dividing equation (2) by (1)

Tsinθ/Tcosθ = qσ/ε₀mg

⇒Tanθ = qσ/ε₀mg

σ = ε₀mg tanθ/q

Now substituting the values of

σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8

σ = (8.85*10^-12 * 0.5774) / 2*10^-8

σ = 5.11*10^-12 / 2*10^-8

σ = 0.255*10^-3 C/m²

7 0
2 years ago
Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
Oksanka [162]

a) The acceleration of the objects is 3.56 m/s^2

b) The tension in the string is 280.8 N

Explanation:

a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

m_1 g - T = m_1 a (1)

where

m_1 = 45.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

T-m_2 g = m_2 a (2)

where

T is the tension in the string

m_2 = 21.0 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

T=m_1g - m_1 a

And substituting into (2),

(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g  = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2

b)

From the previous part of the problem we got an expression for the tension in the string:

T=m_1g - m_1 a

Where we have

m_1 = 45.0 kg

g=9.8 m/s^2

a=3.56 m/s^2 is the acceleration, found in part a)

Susbtituting, we find

T=(45.0)(9.8)-(45.0)(3.56)=280.8 N

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4 0
2 years ago
A FBD of a rocket launching into space should include:
Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up,  the force of gravity pulling it down,    maybe some force of air resistance as the rocket goes fast,   hmmm    Free Body Diagrams  (FBD)  should have any and all forces on the model,  unless they are negligible . or so slight they really make little difference in the total  outcome.  

3 0
1 year ago
Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air
vova2212 [387]
Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
Mass = (0.00118g/cm³)(0.0593 cm³)
Mass = 7 grams of HCN
7 0
2 years ago
You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball
sesenic [268]

Explanation:

The work done equals the change in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work is force times displacement.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

3 0
2 years ago
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