Physics College

can someone help me :(​

Can Someone Help Me :(

Answers

Answer 1

Answer:

1. This is a speed graph.

2. The slope represents how much the item/ thing is moving at how much distance it moved and the minutes it takes it too move how much distance, but the slope is representing the decrease and increase of it.

3. Increasing, kind of rapidly going up since the slope is rising not falling.

Related Questions

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answers

This question is incomplete, the complete question is;

A positive charge of magnitude Q1 = 6.5 nC is located at the origin.

A negative charge Q2 = -3.5 nC is located on the positive x-axis at x = 16.5 cm from the origin. The point P is located y = 10.5 cm above charge Q2.

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answer:

the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C    

Explanation:

Given the data in the question;

Q1 = 6.5 nC, Q2 = -3.5 nC

from the image below, to get our angle ∅

tan∅ = opp/adj

tan∅ = 10.5 / 16.5

tan∅ = 0.636363

∅ = tan⁻¹( 0.636363 )

∅ = 32.47°

also, r1 = √( 16.5² + 10.5²)

r1 = √( 272.25 + 110.25 )

r1 = √382.5

r1 = 19.55 cm = 0.1955 m

Now, the x-component of the electric field at point P due to charge Q1 will be;

Ex = E2cos32.47°

= (kQ1/r1²)cos32.47°

we know that;  k is Coulomb's law constant ( 9 × 10⁹ N.m²/ C²

Q1 = 6.5 nC = 6.5 × 10⁻⁹ C

so we substitute

= ((9 × 10⁹ × 6.5 × 10⁻⁹) / (0.1955)²) cos32.47°

= (58.5 / 0.03822025) × 0.843672

= 1291.33 N/C

Therefore, the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C

during what part of their drive are the not moving?

Answers

C answer is c math ez
The correct answer is c

A car company is doing crash tests and has installed forcesensors in the chairs and seatbelts to measure the forces thatare exerted on crash test dummies during a crash. In oneparticular test the sensors measure a horizontal net force onthe dummy of 39,400 N when the car comes to a complete stopfrom an initial speed of 35 m/s in 1.4 m. Treating the dummyas a point particle and assuming it comes to rest in the samedistance, what is the mass of the test dummy used during thistest

Answers

Answer:

[tex]m=90.1kg[/tex]

Explanation:

From the question we are told that

Force F=39400

Initial speed [tex]V_1=35m/s[/tex]

Distance[tex]d=1.4m[/tex]

Generally the equation for acceleration is mathematically given by

[tex]a=\frac{V^2-U^2}{2d}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=-437.5m/sec^2[/tex]

Generally the equation for mass is mathematically given by

[tex]F=ma\\m=F/a[/tex]

[tex]m=\frac{-39400}{-437.5}[/tex]

[tex]m=90.1kg[/tex]

Besides gravity, what factor keeps the moon and Earth in orbit?

force
inertia
mass
distance

Answers

Answer:

interna

Explanation:

please mark as brainllest

Answer is inertia
Good Luck ;)

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