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3 years ago

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

1 answer:

7 0

Time it takes the projectile to hit the ground after being thrown up:

√h/1/2a

√8/(.5)(9.81)

√8/4.905

√1.630988787

= 1.277101714

= 1. 28

hope this helps :)

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A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of static friction $\mu _s=0.584$

Coefficient of kinetic friction $\mu _k=0.399$

(a) Static friction force is given by $F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N$

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by $F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N$

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2 years ago

A mineral's chemical composition and crystal structure help determine it's ___________.

AnnyKZ [126]

Hiya!

The answer to your question is B.

Physical Properties.

~Hope this helps~

4 0

3 years ago

Change in inherited characteristics over time is called what?

adell [148]

Answer:

characteristics

Explanation:

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6 0

2 years ago

A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

3 years ago

5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co

Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0

2 years ago