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posledela
3 years ago
12

You ride your skateboard to school and tell all your friends you rode at speed of 25 miles per hour

Physics
1 answer:
Ray Of Light [21]3 years ago
5 0
And they think you're cool
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Which term describes when situations, events, or people make demands on your body and mind
mars1129 [50]
<span>Stress
       
Stress is a condition of mental strain or pressure on the body and coming about because of unfavorable or demanding conditions that the person cannot handle at that moment.</span>
6 0
3 years ago
Read 2 more answers
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

t=0.33\ sec

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

5 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
1 year ago
The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a
MariettaO [177]

Answer:

utilization / effects

Explanation:

Utilization equipment are those equipment that makes use of electric energy for the purpose of chemical, electronic, lighting, heating, electro-mechanical or other alike purposes. Hence utilization best suits the first question mark in the question. Secondly, there are associated effects when current flows through a conductor, not responses.

5 0
3 years ago
Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).
JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

6 0
3 years ago
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