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Elena L [17]
3 years ago
6

which of the following values can you find by calculating the slope of a graph of velocity versus time position . displacement .

acceleration . velocity
Physics
1 answer:
statuscvo [17]3 years ago
5 0
The correct answer is: acceleration.

In fact, the slope of the graph of velocity versus time corresponds to the ratio between the variation of v and the variation of t:
\frac{\Delta v}{\Delta t}
But this is exactly the definition of acceleration: a= \frac{\Delta v}{\Delta t}, so acceleration is the correct answer.
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angela uses a force of 25 newtons to lift her grocery bag while doing 50 joules of work. how far did she lift the grocery bags
Evgen [1.6K]
We know that:
W=Fs
50J=25N*s
s=50J/25N
s=2m
5 0
3 years ago
A ball is dropped off a cliff and falls for 24 seconds. How far did the ball fall from the top down
DerKrebs [107]

Answer:

9266 feet

Explanation:

with Earth's gravity and long it fell that's as good as it gets if there was no other factors like wind mass weight but your welcome

4 0
3 years ago
An apparatus similar to the one used in lab uses an oscillating motor at one end to vibrate a long rope with frequency f = 40 Hz
Mice21 [21]

Answer:

The displacement in t = 0,  

y (0) = - 0.18 m

Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

y (0) = 0.25 m *sin ( w*(0) - 225 )

y (0) = 0.25 * -0.707

y (0) = - 0.18 m

5 0
3 years ago
if a piece of sea floor has moved 50 km in 5 million years what is the yearly rate of sea-floor motion?
BartSMP [9]
<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
6 0
3 years ago
Read 2 more answers
Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa
zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, \rho _o = 926 kg/m³

density of the wood, \rho _{wood} = 974 kg/m³

density of water, \rho _w = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0
3 years ago
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