Any winds, like the jet stream, occur because of ____

Calculate the speed of a person who walked 100 m in 25 seconds. Show all of your work. Step 1: Using a light color, highlight th

satela [25.4K]

Answer:

4 m/s

Explanation:

to find m/s, divide the meters term by the seconds term. 100m/25s = 4m/s

4 0

3 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

4 years ago

A value that describes how heavy an object is and is related to the force of gravity is:

Sergio [31]

The weight of an object

6 0

3 years ago

Read 2 more answers

A group of students are provided with three objects all of the same mass and radius. The objects include a solid cylinder, a thi

SOVA2 [1]

Answer:

Sphere, cylinder hoop

Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

In the highest part of the ramp

Em₀ = U = mg y

In the lowest part

Here the energy has part of translation and part of rotation

$E_{mf}$ = $K_{T}$ + $K_{R}$

$E_{mf}$ = ½ m $v_{cm}$ ² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

$v_{cm}$ = w r

w = $v_{cm}$ / r

Let's replace

$E_{mf}$ = ½ I ( $v_{cm}$ / r)²

Energy is conserved

mg y = ½ m $v_{cm}$ ² + ½ I $v_{cm}$ ² / r2

½ (m + I / r²) $v_{cm}$ ² = m g y

½ (1 + I / m r²) $v_{cm}$ ² = g y

$v_{cm}$ = √ [2gy / (1 + I / mr²)]

This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

v = d / t

t = d / v

t = d / (√ [2gy / (1 + I / mr²)])

t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

Sphere is Is = 2/5 M r²

Cylinder Ic = ½ M r²

Hoop Ih = M r²

Let's replace each one and calculate the time

Sphere

ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

ts = (d / √ 2gy) 1.1

Cylinder

tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

tc = (d / √ 2gy) 1.2

Hoop

th = (d / √2gy) √ (1 + mr² / mr²)

th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

th = (d / √ 2gy) 1.41

We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

ts < tc < th

Sphere, cylinder hoop

5 0

3 years ago

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

4 years ago

Any winds, like the jet stream, occur because of _____.