Miners find a rock that was formed from magma deep underground what type of rock could it be

An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (

Leona [35]

Answer:

Explanation:

Given

$v_x(t)=\alpha -\beta t^2$

$\alpha =4\ m/s$

$\beta =2\ m/s^3$

$v_x(t)=4-2t^2$

$v=\frac{\mathrm{d} x}{\mathrm{d} t}$

$\int dx=\int \left ( 4-2t^2\right )dt$

$x=4t-\frac{2}{3}t^3$

acceleration of object

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=-4t$

(b)For maximum positive displacement velocity must be zero at that instant

i.e. $v=0$

$4-2t^2=0$

$t=\pm \sqrt{2}$

substitute the value of t

$x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}$

$x=3.77\ m$

7 0

2 years ago

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s

Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0

1 year ago

The heat capacity of an object depends in part on its ____.

nikdorinn [45]

If I remember it correctly, heat capacity is inversely proportional to mass so the answer is:
The heat capacity of an object depends in part on its a. mass

7 0

2 years ago

A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

7 0

2 years ago

Mark and Balthazar are preparing to conduct neutralization reactions in which they add a base to two different solutions, citric

zhuklara [117]

Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.

5 0

2 years ago

Read 2 more answers