All categories

2 years ago

What does the area of an acceleration time graph represent?

Physics

1 answer:

Leto [7]2 years ago

6 0

The area under the acceleration time graph represents change in velocity

The graph is plotted with acceleration on the vertical axis and time on the horizontal axis. The area under such graph represents change in velocity

You might be interested in

A wave has a frequency of 655 he And is traveling at 455m/sec. what is the wavelength

Vlada [557]

Answer:

v=wavelength ×f

wavelength=v/f=455/655=0.694m

3 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

Un objeto de 1400 g de masa se mueve bajo la acción de una fuerza constante con una aceleración de 0,5 m/s2 , sobre una superfic

igomit [66]

Answer:

if you spoke this in english i can help you out

Explanation:

3 0

3 years ago

An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

Vertical Launch Upwards

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$