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Ksivusya [100]
3 years ago
15

From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h

ave after 4s?
A. 50m / s
B. 60m / s
C. 70m / s
D. 80m / s
Physics
2 answers:
stepan [7]3 years ago
4 0

The answer:

A) 50 m/s

Yuki888 [10]3 years ago
3 0

Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

Find: v

v = at + v₀

v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

v² = (30 m/s)² + (40 m/s)²

v = 50 m/s

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An object with 10 protons 14 electrons and 5 neutrons would have what charge
True [87]

Answer:-4

Explanation:because

10protone +10

14 electrone-14

5 neutrones0 the result will be

10-14=-4

7 0
3 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
3 years ago
Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and
sleet_krkn [62]

Answer:

\alpha=78.4\ rad.s^{-2}

Explanation:

Given:

  • mass of solid cylinder, m=50\ kg
  • diameter of cylinder, d=0.25\ m
  • mass of bucket of water, m_w=20\ kg

<em>When the bucket is released to fall in the well, it fall under the acceleration due to gravity.</em>

We have formula for angular acceleration as:

\alpha=\frac{g}{r}

where:

g = acceleration due to gravity

r = radius of the cylinder

\aplha=\frac{9.8}{0.125}

\alpha=78.4\ rad.s^{-2}

4 0
3 years ago
An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign
dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0
3 years ago
Not a question but need a lot of tips
skad [1K]

DibujaZos da buenos tips e incluso el pinta con colores faber castel son económicos y el dibujo queda epico

8 0
3 years ago
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