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Ugo [173]
3 years ago
6

A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin

g is angled at 60 degree with the roof.
a) find the horizontal distance d it travels. b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?
Physics
1 answer:
Kobotan [32]3 years ago
8 0
<span>c) what is the angle (relative to the horizontal) of the balls initial velocity? </span>
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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
2 years ago
A soccer ball is rolling past Ramon. He kicks it in the same direction that it IS rolling. What will MOST LIKELY happen?
vovikov84 [41]

<u>Answer:</u>

<h2 /><h2>A) The ball will speed up.</h2>
6 0
2 years ago
Read 2 more answers
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
An object accelerates 13.0 m/s2 when a force of 4.4 newtons is applied to it. What is the
velikii [3]

Answer:

0.3384 N

Explanation:

Acceleration = 13 m/s^2

Force = 4.4 N

Force = mass * acceleration

mass = force / acceleration

mass = 4.4 / 13

mass = 0.3384 N

6 0
2 years ago
The Hubble Space Telescope (HST) orbits 569,000m above Earth’s surface. Given that Earth’s mass is 5.97 × 10^24 kg and its radiu
Soloha48 [4]
Refer to the diagram shown below.

M = 5.97 x 10²⁴ kg, mass of the earth
h = 5.69 x 10⁵ m, height of HST above the earth's surface
R = 6.38 x 10⁶ m, radius of the earth

Note that
G = 6.67 x 10⁻¹¹ (N-m²)/kg², gravitational acceleration constant.
R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m

The force between the earth and HST is
F = (GMm)/(R+h)²

Let v = tangential velocity of the HST.

The centripetal force acting on HST is equal to F.
Therefore
m*[v²/(R+h)] = (GMm)(R+h)²
v² = (GM)/(R+h)
    = [(6.67 x 10⁻¹¹ (N-m²)/kg²)*(5.97 x 10²⁴ kg)]/(6.949 x 10⁶ m)
    = 5.7303 x 10⁷ (m/s)²
v = 7.5699 x 103 m/s

Answer
The tangential speed of HST is about 7,570 m/s 

3 0
3 years ago
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