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3 years ago

11. (a)What downward force is acting on you when you go down a waterslide? (b)What type of friction is

Physics

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

a) "gravitation" is the force causing you to go down a waterslide

b) It is "fluid friction" as a solid object (our body) moves over a fluid (the water)

c) It would become "sliding friction" since two solid surfaces slide over each other

d) fluid friction being the weakest friction, switching to sliding friction means a higher decrease in speed and therefore removing the water from a slide will decrease our speed

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A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, $p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s$

Final momentum of the ball, $p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s$

Change in momentum of the ball, $\Delta p=p_f-p_i$

$\Delta p=12-(-12)=24\ kg-m/s$

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0

3 years ago

A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot

Alex_Xolod [135]

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

Distance during acceleration

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

Distance at a constant speed

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

Total distance

Total=27+10.125=37.125 m

3 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0

3 years ago

an astronaut weighs 104 newtons on the moon where the strength of gravity is 1.6 newtons per kilogram what is her mass

Andreas93 [3]

M = W/g
mass (m)
weight (W) and strength of gravity (g)
Therefore the mass of the astronaut is 65 kilograms

7 0

3 years ago

Can a falling object reach terminal velocity in outer space? Explain.

Damm [24]

Ith air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (195 km/h or 122 mph) for a human skydiver. ... (On the Moon, the gravitational acceleration is much less than on Earth, approximately 1.6 m/s2.)

6 0

3 years ago