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3 years ago

WILL GIVE BRAINLIEST AND 24PTS

2 answers:

5 0

A non-moving bike has wheels that aren't spinning and zero angular momentum, which makes it very easy for external torque to change the wheels' direction, making the bike harder to balance. Even when staying relatively motionless, though, a rider can balance a bike with some effort.

Nikolay [14]3 years ago

5 0

Explanation:

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a layer of sandstone is in contact with a mass of granite. the sandstone contains small fragments of the granite. which rock is

lara31 [8.8K]

The granite would be older. As millions of years go by, rocks are affected by weathering and erosion. These processes break down rocks and scatter them. Rocks are broken down into sediments, which mix with other layers, which could have been the reason how the layer of sandstone contains the small fragments of granite.

8 0

3 years ago

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out

anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = \frac{1}{2}mv^{2} + 0$

$2gH = v^{2}$

$v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0\times 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56\times 0.638 = 4.823\ m$

7 0

3 years ago

Read 2 more answers

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

4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

plug in known varibles,

v=2.4*8

v=19.2m/s

4 0

2 years ago

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, $\omega=33\frac{1}{3} rev/min$ , radius of the record $r=0.1m$ and the distance between any two successive bumps on the groove as $d=1.75mm$ .

The linear speed of the record in meters per second is:

$v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\$

#From $v$ above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

$Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199$

Hence the bumps hit the stylus at around 199hit/s

8 0

3 years ago