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marissa [1.9K]
3 years ago
9

A ball is kicked at an angle of 35 degrees with the ground. What should be the initial velocity of the ball so that it hits a ta

rget that is 30 meters away at a height of 1.8 meters?
Physics
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

18.5 M/S

Explanation:

trust me

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Figure 2. sketch of bar and magnetic field lines observations PLEASE HELPPPP ​
labwork [276]
Hope this sketch helps

5 0
2 years ago
Imagine a 10kg block moving with a velocity of 20m/s to the left.
lara31 [8.8K]
Kinetic energy , KE= [1/2]m*v^2

m = 10 kg
v=20m/s

KE = [1/][(10kg)(20m/s)^2 = [1/2](10kg)(400m^2/s^2) = 2000 joule

Answer: 2000 joule
3 0
3 years ago
A biker is pedaling at a constant speed of 36 km/h. During the last 10 s of the race, he increases his speed with a constant acc
adell [148]

Answer:

54 km/h

Explanation:

given,

speed of the biker = 36 Km/h

time = 10 s

acceleration = 0.5 m/s²

speed at which it crosses the finish line  = ?

v = 36 x 0.278 = 10 m/s

using equation of motion

v = u + a t

v = 10 + 0.5 x 10

v = 15 m/s

v = 15 x 3.6 = 54 km/hr

speed at which the biker crosses the finish line is equal to 54 km/h

4 0
3 years ago
A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do
Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0
2 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
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