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3 years ago

Please help! Will give brainliest. 10 points. Show work!

Physics

1 answer:

Natasha_Volkova [10]3 years ago

5 0

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

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Then the girl is very strong. 10kg bags are really heavy

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3 years ago

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

5 0

3 years ago

What is the maximum powr of a module in Watts to the nearest whole Watt?

KonstantinChe [14]

Complete question is;

You are looking at a module specification

sheet that has the table of information

below. What is the maximum power of this

module in Watts to the nearest whole Watt?

Value

Polycrystalline si

Characteristic

Cell Type

Cell

Configuration

Voc

160 in series

137.2 V

V_imp: 29.3 V

Ilsc: 8.60 A

I_Imp: 8.02 A

Dimensions (mm/in): 1000 x 1600 x 50 mm / 39.4" x 63" x 2"

Weight: 10 kg / 22 lbs

Answer:

P ≈ 235 Watts

Explanation:

Formula for power is;

P = IV

Now, for maximum power, we will make use of I_imp and V_imp given

Thus, P = I_imp × V_imp

We are given;

V_imp: 29.3 V

I_Imp: 8.02 A

Thus: P = 8.02 × 29.3 = 234.986 Watts

We are to approximate to the nearest whole watt.

Thus: P ≈ 235 Watts

6 0

3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

3 years ago