Ask question

Ask question

All categories

Thepotemich [5.8K]

3 years ago

6

A ball is launched at 24m/s at a 45 degree angle as shown. What is the horizontal component of the velocity (Vx)? (don't include

units)

1 answer:

zaharov [31]3 years ago

5 0

For the horizontal component <u>we have:</u>

Vx = Vo * cos α

Replacing we have:

Vx = 24 * cos 45°

Resolving:

Vx = 24 * 0.525...

Vx = 12.607

The horizontal velocity is <u>12,607 m/s.</u>

You might be interested in

answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in

Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0

3 years ago

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago

Which terms describe space objects that are small chunks of rock and debris that are smaller than 1 km?

UkoKoshka [18]

I would say that it is Meteors, Meteorites, and comets. (METEORS may be wrong)

5 0

3 years ago

Read 2 more answers

Defenition of takes place in body cells

swat32

Answer:

the nucleus-containing central part of a neuron exclusive of its axons and dendrites that is the major structural element of the gray matter of the brain and spinal cord, the ganglia, and the retina. — called also perikaryon, soma.

6 0

4 years ago

Read 2 more answers

A 2.3 kg , 20-cm-diameter turntable rotates at 110 rpm on frictionless bearings. Two 460 g blocks fall from above, hit the turnt

boyakko [2]

Answer:

The correct solution is "64 RPM".

Explanation:

The given values are:

Mass,

M = 2.3 kg

Diameter,

D = 20 cm

i.e.,

= 0.2 m

Rotates at,

N = 110 rpm

Mass of block,

m = 460 g

i.e.,

= 0.46 kg

According to angular momentum's conservation,

⇒ $I_1\omega_1=I_2\omega_2$

then,

⇒ $I_1=\frac{1}{2}MR_2$

On substituting the values, we get

⇒ $=\frac{1}{2}\times 2.3\times (0.1)^2$

⇒ $=\frac{1}{2}\times 0.023$

⇒ $=0.0115 \ kg \ m^2$

Now,

⇒ $I_2=I_1+2mR^2$

$=0.0115+2\times 0.46\times (0.1)^2$

$=0.0115+0.0092$

$=0.02 \ kg \ m^2$

then,

⇒ $0.0115\times 110=0.02\omega_2$

⇒ $1.265=0.02\omega_2$

⇒ $\omega_2=\frac{1.265}{0.02}$

⇒ $=63.25 \ or \ 64 \ RPM$

7 0

3 years ago