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CaHeK987 [17]
3 years ago
15

Your friend asks you for a glass of water and you bring her 5 milliliters of water. Is this more or less than what she was proba

bly expecting? Explain your reasoning.
Physics
1 answer:
Eddi Din [679]3 years ago
3 0
Probably not what you were expecting... the average bottle of water is 24 ounces. 5 milliliters is about the amount of water in a spoon. Hope this helps!!!
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If Sally is standing on a 200m tall cliff and throws a ball at 40m/s at a 30° angle to the horizontal: a. What is the ball's ini
Irina-Kira [14]

Answer:

(a) 20 m/s (j) m/s

(b) 20√3 m/s (i) m/s

(c) 2.04 s

(d) 20.4 m

Explanation:

In order to solve the problem, you have to apply the <em>Projectil Motion</em> equations.

For part (a) and (b) you have to obtain the components of the initial velocity vector. The direction forms a 30° angle to the horizontal and the modulus (speed) was given. Therefore:

Applying trigonometric identities (Because the initial velocity is the hypotenuse of a right triangle with angle 30° to the horizontal)

Vx: 40Cos(30°)=20√3 m/s

Vy:40Sin(30°)= 20 m/s

The initial velocity in the y direction is: 20 m/s (j) m/s

The initial velocity in the x direction is: 20√3 m/s (i) m/s

Where i and j are the unit vectors.

For part (c) you have to apply the following vertical motion equation:

Vy=Voy-gt

where Voy is the initial velocity, g is gravity and t is the time

The ball reaches its max height when Vy=0 therefore:

0=Voy-gt

Solving for t:

t=Voy/g=20/9.8= 2.04 seconds

For part (d) you have to apply the other vertical motion equation which is:

y=yo+Voyt-0.5gt²

Where yo is the initial position.

Replacing t=2.04 s, yo=0 m, Voy=20 m/s and solving for y:

y=0+(20)(2.04)-(0.5)(9.8)(2.04)²

y=20.4 m

3 0
3 years ago
A constant force of friction 50N is acting on a body of mass 200 Kg moving initially with a speed of 15 m/s. How long does the b
Zina [86]

Answer:

F=-50N

M=200kg

U=15m/s

F=Ma

a=F/M=-50/200=-0.25

V^2-U^2=2aS

0-(15)^2=2(-0.25)S

S=-225/-0.5=450m

V=U+at

0=15-0.25t

t=-15/-0.25=60s

Explanation:

Hope this helps, let me know if you have any questions!

Have a great day.

8 0
4 years ago
You take a trip that covers 240 kilometers and takes 4 hours. Your average speed is
Lorico [155]

Answer:

Your average speed is 60 kilometers per hour

Explanation:

240 divided by 4 is 60

240/4=60

7 0
3 years ago
A source injects an electron of speed v = 2.9 × 107 m/s into a uniform magnetic field of magnitude B = 1.7 × 10-3 T. The velocit
34kurt

Answer:

r = 0.664 m.

Explanation:

Let's write the equation of the magnetic force, the blacks syndicate vectors

       F = q v x B

From this expression we see that the force is perpendicular to the velocity and the field, so it is a centripetal force, the modulus of the force is

      F = q v B sinT

We write Newton's second law

      F = m a

      a = v² / r

     q v B sinT = m v² / r

     r = m v / (q B sinT)

Let's calculate

     r = 9.1 10-31 2.9 107 / (1.6 10-19 1.7 10-3 sin8.4)

     r = 26.4 10-24 / 0.3973 10-22

     r = 0.664 m

This is the distance from where the electron penetrates

7 0
4 years ago
The Sun radiates energy at a rate of about 4×1026W. Earth is about 150×106km from the Sun.
sergiy2304 [10]

The average intensity received on the Earth's surface is 12.6*10^{10} W/m^{2}

​Intensity = Rate  of  energy radiated/4\pi R^{2}

where  R is the distance of the earth from the sun.

I=4*10^{26}/4\pi *[150*106*10^{3}]^{2}

I=0.12597*10^{12}

I=12.6*10^{10} W/m^{2}

What is the meaning of the intensity of sunlight?

Sun intensity refers to the amount of incoming solar energy, or radiation, that reaches the Earth's surface. The angle at which the rays from the sun hit the Earth determines this intensity.

What is the average intensity received on the Earth's surface?

The average intensity received on the Earth's surface is given by:

I=rate of energy radiated/4\pi R^{2}

Thus, the average intensity received on the Earth's surface is 12.6*10^{10} W/m^{2}

To know more about the intensity received from the sun:

brainly.com/question/29579262

#SPJ1

8 0
1 year ago
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