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professor190 [17]
3 years ago
9

You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th

e acceleration of the dragster. Would you use a. a) Linear equation
b) Quadratic equation
c) cubic equation
d) a higher order equation
Physics
1 answer:
Ivenika [448]3 years ago
8 0

Answer:

a) Linear equation

Explanation:

Definition of acceleration

a=\frac{dv}{dt}\\

if a=constant and we integrate the last equation

v(t)=v_{o}+a*t

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

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Answer:

inertia

Explanation:

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When is thermal equilibrium achieved between two identical objects?
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The answer is C hope this helps <span />
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Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
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0 0
3 years ago
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
a 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. if the c
velikii [3]

consider the velocity towards the pitcher as positive

m = mass of the baseball = 0.145 kg

v₀ = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = 3 x 10⁻³ sec

F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

3 0
3 years ago
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