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4 years ago

PLS HELPPPP WILL AWARD BRAINLIEST AND POINTS

Physics

1 answer:

Yuri [45]4 years ago

7 0

Answer:

Here

Length (l) = 6 m

height (h) = 3 m

Load(L) = 500 N

Effort (E) = ?

we know the principal that

E * l = L * h

6 E = 500 * 3

6E = 1500

E = 250

therefore 250 N work is done on the barrel.

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Kara built a model of how molecules behave when they are in solid form. She filled a clear plastic box with marbles to the very

marishachu [46]

Answer:

The marbles must be held in a box, but solids hold together without a container.

Explanation:

study island!!!

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4 years ago

g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the drag

melisa1 [442]

The dragster's velocity v at time t with constant acceleration a is

$v=at$

since it starts at rest.

After 2.1 s, it will attain a velocity of

$v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)$

or 92.4 m/s.

Doubling the time would double the final velocity,

$v=a(2t)=2at$

so the velocity would be twice the previous one, 184.8 m/s.

The dragster undergoes a displacement x after time t with acceleration a of

$x=\dfrac12at^2$

if we take the starting line to be the origin.

After 2.1 s, it will have moved

$x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2$

or 88 m.

Doubling the time has the effect of quadrupling the displacement, since

$x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)$

so after 4.2 s it will have moved 352 m.

5 0

3 years ago

Please help me

iVinArrow [24]

Answer:

0.368 Newtons

Explanation:

3 0

3 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

Stephen Strasburg throws out the opening pitch at Nationals Park. The ball starts at rest, but

tankabanditka [31]

Answer:

Force = 304.5 Newton

Explanation:

Given the following data;

Initial velocity, u = 0m/s (since it started from rest).

Final velocity, v = 42m/s

Time, t = 0.02 secs

Mass = 0.145 kg

To find the force;

First of all, we would have to determine the acceleration of the ball;

Acceleration = (v - u)/t

Acceleration = (42 - 0)/0.02

Acceleration = 42/0.02

Acceleration = 2100 m/s²

Now, we solve for the force;

Force = mass * acceleration

Force = 0.145 * 2100

Force = 304.5 Newton

4 0

3 years ago