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hoa
20 days ago
11

Write a hypothesis about the effect of the racetrack’s height on the speed of the car using the format of "if…then…because….” Be

sure to answer the lesson question "How can motion be described?” 
Physics
2 answers:
Ostrovityanka [2.2K]20 days ago
4 0
<span>If the length of the racetrack measures 36 inches and is set at heights of 8 inches, 16 inches, and 24 inches, with the toy car rolling down in times of 7.8 seconds, 7.3 seconds, and 6.6 seconds respectively, this indicates that the height of the racetrack influences the speed of the toy car. The increase in height results in a quicker descent due to the effect of gravity on velocity, which describes the rate of motion. This scenario illustrates motion as a concept representing movement or a position change given certain circumstances over time.</span>
kicyunya [2.2K]20 days ago
3 0

As the height of a sloped racetrack is raised, the velocity of a toy car moving on the track will also increase due to greater acceleration.

(I just did this) 
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Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.
Sav [2226]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

8 0
1 month ago
Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x
ValentinkaMS [2425]

Response:

210.3 degrees

Justification:

The total force acting on charge A is 59.5 N

Apply the x and y components of the net force to determine the direction

atan (y/x)
8 0
1 day ago
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Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and
Softa [2029]
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26

For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
3 0
27 days ago
Read 2 more answers
The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that
Maru [2355]

Answer:

The correct option is 80 dB.

Explanation:

The transformation of sound intensity level into sound intensity utilizes the formula

[D] = 10 log (I/I₀)

Where I₀ = 10⁻¹² W/m²

[D] results in 100 dB

100 = 10 log (I/I₀)

Log (I/I₀) converts to 10

(I/I₀) = 10¹⁰

I is determined as I₀ × 10¹⁰ = 10⁻¹² × 10¹⁰ = 10⁻² W/m²

Sound intensity inversely relates to the square of the distance from the source.

I ∝ (1/d²)

I can be expressed as k/d²

When d = 1 m, the intensity is 10⁻² W/m²

Thus, 0.01 = k/1

Providing that k = 0.01 W

For d = 20 m, we can calculate I

I = 0.01/20² = 2.5 × 10⁻⁵ W/m²

With four neighbors mowing their lawns concurrently,

I = 4 × 2.5 × 10⁻⁵ = 10⁻⁴ W/m⁻²

The sound intensity level in decibels is represented as

[D] = 10 log (I/I₀)

[D] = 10 log (10⁻⁴/10⁻¹²)

[D] = 10 log (10⁸)

[D] = 10 × 8 = 80dB

4 0
18 days ago
Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm
Sav [2226]

In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

frequency = 1 / wavelength

Calculating frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

 

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

7 0
7 days ago
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