Draw vectors to represent these motions:<br>moving 3 cm to the right<br>moving 2 cm to the left

Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run?

Vedmedyk [2.9K]

Speed is equal to distance traveled divided by the time. So it's 3.5 m/s

6 0

3 years ago

50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui

soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0

3 years ago

when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why

Lady_Fox [76]

Answer:

hsvshxansjusjsnwjwisks

4 0

3 years ago

Read 2 more answers

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0

3 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

Draw vectors to represent these motions:moving 3 cm to the rightmoving 2 cm to the left