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postnew [5]
3 years ago
9

Barry is conducting an experiment and rolls a tennis ball down a ramp. Which best describes the motion of the tennis ball? It do

es not exhibit projectile motion and follows a straight path down the ramp. It does not exhibit projectile motion and follows a parabolic path down the ramp. It exhibits projectile motion and follows a straight path down the ramp. It exhibits projectile motion and follows a parabolic path down the ramp.
Physics
2 answers:
yawa3891 [41]3 years ago
7 0

Answer:

A. It does not exhibit projectile motion and follows a straight path down the ramp.

puteri [66]3 years ago
4 0

Answer:

It does not exhibit projectile motion and follows a straight path down the ramp.

Explanation:

Projectile motion is the motion of the object that is projected into air. The only force acting on it must be the earth's gravity. The object projected into air is called the projectile and its path is called the trajectory.

In our case Barry is not throwing the tennis ball into air. It is just moving along the surface. It is just like any other movement on a flat surface The only difference it is happening at an angle.

In our case there is friction acting on the tennis ball, but only earth's gravity is supposed to act on the tennis ball.

So it does not exhibit projectile motion and follows a straight line path down the ramp

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Mass of water= 357g density water= 1.0g/cm3
aliya0001 [1]

m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3

8 0
3 years ago
Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given tha
pshichka [43]
For resistance we have R=ρ l/a
 thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ

σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
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6 0
3 years ago
Read 2 more answers
Baseball player swings and hits a pop fly straight up in the air to the catcher. the height of the baseball in meters t seconds
andre [41]
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1

When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s

Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.

The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
          = 60.903 m

Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)

4 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height
Ahat [919]

Answer:

V_{3.01}=-93.2m/s

V_{3.005}=-93.1m/s

V_{3.002}=-93.04m/s

V_{3.001}=-93.02m/s

V_{3}=-93m/s

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

Y_{3}=-99m/s

Y_{3.01}=-99.932m/s

Y_{3.005}=-99.4655m/s

Y_{3.002}=-99.18608m/s

Y_{3.001}=-99.09302m/s

Replacing these values into the formula for average velocity:

V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s

V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s

V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s

V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s

To know the actual velocity, we derive the position and we get:

V=27-40t = -93m/s

5 0
3 years ago
A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.
Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0
3 years ago
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