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Verdich [7]
2 years ago
9

Finding the area of a trapezoid on a velocity versus time graph will tell you

Physics
1 answer:
julsineya [31]2 years ago
6 0
The answer is c. velocity
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Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
nalin [4]

Answer:

t = 1.02 s

Explanation:

The computation of the time required is shown below:

The package speed for belt is

= 3 -  1

= 2 m/s

Moreover, the decelerative force would be acted on the block i.e u.m.g

So, the decelerative produced

= 0.2 × 9.81

= 1.962 m/s^2

And, final velocity = 0

v = u - at

here

V = 0 = final velocity

 u = 2 m/s

so,

0 = 2 - 1.962 × t

t = 1.02 s

5 0
3 years ago
An object is accelerating if there is a change in speed and/or ________.
Alexxx [7]
A change in C. Direction.
6 0
3 years ago
Read 2 more answers
Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a
Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

Initial Velocity of ball V_i,b = 0

Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b   ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p)   ...2

The distance that the platform moves d:

d = V_f,p *t  ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p)   .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

The platform moves 0.05312 m to the opposite to which the ball is thrown.

5 0
3 years ago
Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima
Svet_ta [14]

Answer:

5.241\ \text{m/s}

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = 9.81\ \text{m/s}^2

h = Altura = 1.4\ \text{m}

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}

La velocidad máxima que puede alcanzar el coche es 5.241\ \text{m/s}.

8 0
3 years ago
The 1811-1812 new madrid earthquakes were different from most california earthquakes because
g100num [7]

Answer:

New Madrid earthquakes are centered in a plate interior, meaning they were intraplate earthquakes.

Explanation:

The 1811-1812 new madrid earthquakes were different from most california earthquakes because the New Madrid earthquakes were centered in a North American plate interior caused by stress within the plates.

The New Madrid earthquakes often referred to as the "Tecumseh's Comet" was the biggest earthquake in America history.

4 0
3 years ago
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