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VLD [36.1K]
3 years ago
14

A boat is moving at a rate of 15.0 meters per second. Its speed is then decreased uniformly to 3.0 meters per second. It takes 4

seconds for the speed to decrease. What is the magnitude of the deceleration of the car?
Physics
1 answer:
marin [14]3 years ago
5 0

Answer:

3m/s

Explanation:

15m/s - 3m/s

12/4

3m/s

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How can a tennis ball and a bowling ball have the same momentum ?
Verdich [7]
C would be the right answer edu
7 0
3 years ago
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the
sashaice [31]

Answer:

52.47706 mph

54.5 mph

Explanation:

The average speed is given by

V_{av}=\dfrac{Distance}{Time}

V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph

Julie's average speed on the way to Grandmother's house is 52.47706 mph

V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph

Average speed on the return trip is 54.5 mph

4 0
3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
Alex777 [14]

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

7 0
3 years ago
Which formula can be used to find the tangential speed of an orbiting object?
Luden [163]
The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.
6 0
3 years ago
Read 2 more answers
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.
tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

4 0
3 years ago
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