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3 years ago

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A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed v0. At the same time, a s

econd student drops a lighter blue ball from the same balcony. Neglecting air resistance, which statement is true? The blue ball reaches the ground first. The balls reach the ground at the same instant. The red ball reaches the ground first. Both balls hit the ground with the same speed. None of the above statements is true. Incorrect: Your answer is incorrect.

1 answer:

liraira [26]3 years ago

6 0

Without air resistance, both balls reach the ground at the same instant.

Neither horizontal motion nor weight affects vertical motion.

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How long does it take the lava bomb to reach its maximum height? Answer with three significant digits and the correct unit. A sm

wlad13 [49]

Answer:

The time taken to reach the maximum height is 3.20 seconds

Explanation:

The given parameters are;

The initial height from which the volcano erupts the lava bomb = 64.4 m

The initial upward velocity of the lava bomb = 31.4 m/s

The acceleration due to gravity, g = 9.8 m/s²

The time it takes the lava bomb to reach its maximum height, t, is given by the following kinematic equation as follows;

v = u - g·t

Where;

v = The final velocity = 0 m/s at maximum height

u = The initial velocity = 31.4 m/s

g = The acceleration due to gravity = 9.8 m/s²

t = The time taken to reach the maximum height

Substituting the values gives;

0 = 31.4 - 9.8 × t

∴ 31.4 = 9.8 × t

t = 31.4/9.8 ≈ 3.204

The time taken to reach the maximum height rounded to three significant figures = t ≈ 3.20 seconds

4 0

3 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

3 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm

mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

$f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}$

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

$V_{sph}=\frac{4\pi}{3}r^3$

So the diameter is:

$D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m$

6 0

2 years ago

If youre going 70 miles an hour for 70 miles, how long would it take you to travel those 70 miles

Anna11 [10]

490 miles in total ur welcome

6 0

3 years ago