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3 years ago

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a net force f accelerates a mass m with an acceleration a. if the same net force is applied to mass 5m, then the acceleration wi

ll be

1 answer:

diamong [38]3 years ago

3 0

Answer:

F=ma

a =F/m

a =F/5m ms-2,,,,,

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The eye of a hurricane passes over grand bahama island in a direction 60.0Â° north of west with a speed of 41.0 km/h. three hour

Ksivusya [100]

initial velocity is given as 41 km/h at 60 degree North of West

$v_i = -41 cos60\hat i + 41 sin60 \hat j$

$v_i = -20.5 \hat i + 35.5 \hat j$

After some time the velocity is given as

$v_f = 25 \hat j$

now we can find the acceleration

$a = \frac{v_f - v_i}{t}$

$a = \frac{25\hat j - (-20.5 \hat i + 35.5 \hat j)}{3}$

$a = 6.83 \hat i - 3.5 \hat j$

now the distance is given by

$d = v*t + \frac{1}{2} at^2$

$d = (-20.5 \hat i + 35.5 \hat j)*4.5 + 0.5*(6.83 \hat i - 1.5 \hat j)* 4.5^2$

$d = -92.25\hat i + 159.75\hat j + 69.15\hat i - 15.19\hat j$

$d = -23.1 \hat i + 144.56\hat j$

so the magnitude of distance is

$d = \sqrt{23.1^2 + 144.56^2} = 146.4 km$

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3 years ago

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Coulomb’s law is similar to what other law in physics?

Yuki888 [10]

Newton's law of gravitation

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3 years ago

10 kg cart and a 5 kg cart are placed on identical surfaces. The 10 kg cart experiences a net force of 12 N to the left, while t

SashulF [63]

F=ma

For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left

For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left

Therefore, the lighter (5kg) cart experiences a greater acceleration.

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3 years ago

A secant and a tangent meet at a 90Ă‚Â° angle outside the circle. What must be the difference between the measures of the interc

kozerog [31]

The difference between the measures of the intercepted arcs of the given circle is 45°.

A circle is defined as the set of points in a plane equidistant to each, such that it forms a closed two-dimensional figure, which is known as a circle.

A line intersecting a circle at a minimum of two distinct points is known as a secant and the line touching the circle at only one point, is known as the tangent of a circle.
The intercepted arc is the section of the circumference of a circle such that it is either encased by two chords or a line segment, meeting at a single point.

We are given that the angle subtended by secant and tangent, outside the circle is,

$\theta =90^{\circ}$

And angle measured inside the circle is,

$\phi = \theta /2\\\\\phi = 90/2\\\\\phi = 45{^\circ}$

So, the difference between the measures of the intercepted arcs is,

$\theta' = \theta - \phi\\\\\theta' = 90-45\\\\\theta' =45^{\circ}$

Thus, we can conclude that the difference between the measures of the intercepted arcs of the given circle is 45°.

learn more about the tangent here:

brainly.com/question/14022348

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3 years ago

A horizontal pipe carrying water tapers from 200mm diameter

Tomtit [17]

Answer:

Delta pressure = 45938[Pa]

Explanation:

This is a classic bernoulli equation problem, but first we must find the velocity at both ends of the pipe, knowing the diameters of each end.

$Q=V*\frac{\pi*d^{2} }{4} \\where:\\V = velocity[m/s]\\d = diameter [m]\\Q = volumetric flow[m^{3}/s]$

$Q=55[\frac{Lt}{s}] * \frac{1m^3}{1000Lt}\\Q=0.055[\frac{m^3}{s}]$

Now we can calculate the two velocites VA and VB

Using the bernoulli equation we have:

$P_{A}+ro*V_{A}^{2} +h_{A}*g*ro = P_{B}+ro*V_{B}^{2} +h_{B}*g*ro \\as we have the same elevation\\h_{A}*g*ro = h_{B}*g*ro\\Therefore\\P_{A}+ro*V_{A}^{2} = P_{B}+ro*V_{B}^{2}\\P_{B}- P_{A}= ro*V_{A}^{2} - ro*V_{B}^{2}\\\\P_{B}- P_{A}=ro*(V_{A}^{2} - V_{B}^{2})$

Where ro = density of water = 1000[kg/m^3]

$P_{B}-P_{A}= 1000*(7^2-1.75^2)\\P_{B}-P_{A}= 45937.5[Pa]$

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3 years ago