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3 years ago

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An object with a mass of 5.0 kg accelerates 2.8 m/s2 towards right when an unknown force is applied to it. What is the amount of

the force? What is the direction of force?

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

4n

Explanation:

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A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac

lana66690 [7]

Answer:

charges of the beads is 1.173 × $10^{-15}$ C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know force for mass

force = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 × $10^{9}$ )

so charges of the beads = 1.173 × $10^{-15}$ C

3 0

2 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

2 years ago

Which of the following distinguishes the isotope uranium-238 from the isotope uranium-235?

WITCHER [35]

Answer: A

Explanation:

Isotopes of different elements differ by the number of neutrons inside the nucleus.

8 0

2 years ago

Read 2 more answers

A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for

luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

$\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}$

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

$\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}$

In conclusion, the normal force acting on the block is 50 N.

5 0

2 years ago

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8

hammer [34]

Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
$W= \frac{1}{2} k (\Delta x)^2 
$
where k is the elastic constant of the spring and $\Delta x$ is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
$W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J$

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
$W_W = -F_{//} (x_2 -x_1)$
where the negative sign is given by the fact that $F_{//}$ points in the opposite direction of the displacement of the cart, and where
$F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N$
therefore, the work done by the weight is
$W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J$

8 0

2 years ago