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3 years ago

What equation can be used to solve for acceleration

1 answer:

3 0

Answer: acceleration can be calculated when a known force is acting on an object of known mass. Newton’s law can be represented by the equation F net = m x a, where F net is the total force acting on the object, m is the object’s mass, and a is the acceleration of the object.

Explanation:

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This engineer invented the iron wire to replace hemp rope.

IceJOKER [234]

The answer is A John Roebling.

7 0

4 years ago

Read 2 more answers

You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro

ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

5 0

3 years ago

A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of

kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0

3 years ago

A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

6 0

3 years ago

In figure 1, charge q2 experiences no net electric force. What is q1?

lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

$F = k\frac{q_1*q_2}{r^2}$

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

$F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}$

And we know that it must be equal to zero, so we can write it as:

$F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\$

The parenthesis must be equal to zero, so we can write:

$\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0$

And now we can solve this for q₁ to get:

$q_1 = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC$

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0

3 years ago