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2 years ago

I really need help with the graphs

Physics

1 answer:

nata0808 [166]2 years ago

3 0

Answer:

t = 2s

Explanation:

When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)

Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.

With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.

So at t = 2s, we have an instantaneous acceleration of -2m/s².

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Having just enough weight to achieve all three states of buoyancy with only minor adjustments in the water is the definition of:

eduard

Answer:

Proper weighting

Explanation:

Proper weighing involves the condition of a scuba diver that is fully geared having a near empty tank and the BCD emptied with a held breadth is expected to float at eye level

The fundamental of adequate or good buoyancy of a scuba diver is to ensure proper weighting when diving, With proper weighting, there is more control for the diver when a safety stop is required. There is less need to carry excess weight that increases drag and gas consumption.

6 0

3 years ago

I’d really appreciate it if someone could help

Vanyuwa [196]

Answer:

0.25

Explanation:

Friction force is the normal force times the coefficient of friction.

F = Nμ

50 N = (200 N) μ

μ = 0.25

6 0

3 years ago

When Earth pulls on an object, that object also pulls on Earth. The values of these two forces are . This phenomenon can be expl

algol13

The values of these two forces are equal. Your weight on Earth is equal to the Earth's weight on you. When you and the Earth fall toward each other, your acceleration is greater than the Earth's acceleration, because your mass is less than the Earth's mass.

5 0

3 years ago

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Calculate V1 and V2 ( V = voltage )

Kitty [74]

FORMULA:

V = IR, where V = P.D; I = Current; R = Resistance.

ANSWER:

Total equivalent resistance for circuit:

R(eq) = R1 + R2 [It is in series]

330Ω + 470Ω
800Ω

Now, Current passing through whole circuit:

I = V/R

16/800
1/50 ampere

We know that, In series combination current passing through whole circuit is same.

So, V¹ = IR¹

V¹ = 1/50 × 330

V¹ = 6.6 volt

And V² = IR²

V² = 1/50 × 470

V² = 9.4 volt

7 0

3 years ago

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