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cupoosta [38]
3 years ago
6

Please helppp

Physics
1 answer:
Y_Kistochka [10]3 years ago
6 0

Answer:

F = 17.3 kN

Explanation:

The normal force must support the weight of the car plus provide for the needed centripetal acceleration.

F = m(g + v²/R ) = 1000(9.8 + 15²/30) = 17,300

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Kyle has a mass of 54kg and is moving at 3 m/s what is his kinetic energy
Mars2501 [29]

Answer:

243J

Explanation:

K.E = 1/2 x 54 x 3^2

K.E = 1/2 x 54 x 9

K.E = 1/2 x 486

K.E = 486/2

K.E = 243J

6 0
3 years ago
A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force or Tour
Tanzania [10]

Answer:.

Explanation:

7 0
3 years ago
Why is Pluto now called a dwarf planet
zvonat [6]
Before Pluto was discovered, it was predicted. Astronomers had observed that massive objects can affect the orbits of its neighbors, and, after seeing deviations in the orbits of Uranus and Neptune, assumed something substantial existed beyond their orbits.
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The International Astronomical Union decided that full-sized planets must orbit the sun, have a round shape, and have cleared their orbits of other objects. Pluto fulfills the first two criteria, but not the third.
It still goes around the sun, it’s round enough, it’s got moons, and behaves like a planet, but the idea is that Pluto did not form the same way as the rest of the planets. Pluto’s orbit is both eccentric and inclined more than the rest of the planets by about 17 degrees. That’s suggests something is different about this object.
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7 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base
riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of  ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

h=vt+\dfrac{1}{2}gt^2

v= 0 m/s

h=\dfrac{1}{2}gt^2

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0
3 years ago
Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex
AURORKA [14]

Answer:

554.27N

Explanation:

(a)  The max frictional force exerted horizontally on the crate and the floor is,

Substitute the values,

μs=0.5

mass=113kg

g=9.81m/s

Ff=μsN

   =μsmg

   =(0.5 x 113 x 9.81)

Ff=554.27N

3 0
3 years ago
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