Quit my nose so you have a bad guy really good guy

Who is wwe universal champion

mina [271]

Answer:

Roman reigns

Explanation:

is the wwe universal hampion

7 0

3 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

$a=4.2 m/s^2$

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

$v=0+(4.2)(6.5)=27.3 m/s$

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

$d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m$

Which is larger than 30 m: this means that the gazelle covers the 30 m during its accelerated motion. Therefore, we can use again the equation:

$d=\frac{1}{2}at^2$

And substituting d = 30 m, we find the time:

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s$

3. 10.6 s

In this case, the distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

$d' = 200-88.7 =111.3 m$

Therefore, the time taken to cover this distance is

$t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s$

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s

6 0

4 years ago

A projectile is fired over horizontal ground from the origin with an initial speed of 60 m/s. What firing angles will produce a

horsena [70]

Answer:

Angle θ ≅ 21.5°

Explanation:

Given:

speed Vi= 60 m/s, Range R= 250 m, g=9.81 m/s²

To find:

Angle θ = ?

Sol: we know that Rang R = Vi² sin 2θ / g

⇒ Sin 2θ = g×R / Vi²

Sin 2θ = (9.81 m/s² × 250 m) / ( 60 m/s)²

Sin 2θ = 0.68125

2θ = Sin ⁻¹ (0.68125)

2θ = 42.9413

θ = 42.9413 / 2

θ = 21.4706 ≅ 21.5°

8 0

3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$