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laila [671]
3 years ago
8

What is the mass of an object moving with a velocity of 4 m/s and a kinetic energy of 2000 Joules?

Physics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:

m = 250 kg

Explanation:

Hope this helps

ki77a [65]3 years ago
3 0

Answer:

250 kg

Explanation:

see pic

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what is formed when two or more substances are so evenly mixed that you can't see the different parts
Andre45 [30]
A homogenous mixture
5 0
4 years ago
A uniformly charged, thin ring has radius 15.0 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance
djyliett [7]

Answer:

A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.

B. Veloctiy (Vb) = 1.66m/s

Explanation:

Given the following data

x(a) = 0.3m

x(b) = 0

q = 1.6×10^-19

Q = 24nc

r = 0.15m

Required: the motion of the electron and the velocity (Vb)

1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B

2. Potential energy and kinetic energy are given by

U(a) + K(a) = U(b) + K(b). . .1

Initial P.E and K.E are given as

U(a) = kQ/√x²(a) + a2

By substitution, we have

U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)

U(a) = -1.03×10^-16

Final P.E and K.E are given as

U(b) = KQ/√x²(b) + a2

By substitution, we have

U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²

U(b) = -2.3×10^16

3. By substitution into equation 1 becomes

-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2

V(b) = √2×1.27×10^-16/9.1×10^31

V(b) = 1.66×10^7m/s

4 0
4 years ago
A pair of closely spaced slits is illuminated with 650.0-nm light in a Young's double-slit experiment. During the experiment, on
Masteriza [31]

Answer:

The minimum thickness of the Lucite plate is 0.670 μm.

Explanation:

Given that,

Wavelength = 650.0 nm

Index of refraction = 1.485

We need to calculate the minimum thickness of the Lucite plate

Using formula of thickness

t(n-1)=\dfrac{\lambda}{2}

t=\dfrac{\lambda}{2(n-1)}

Where, n = Index of refraction

\lambda = wavelength

Put the value into the formula

t=\dfrac{650.0\times10^{-9}}{2(1.485-1)}

t =0.670\ \mu m

Hence, The minimum thickness of the Lucite plate is 0.670 μm.

7 0
3 years ago
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Romashka-Z-Leto [24]

3-squared + 4-squared = 5-squared

The bug's displacement is 5cm east.


4 0
4 years ago
Read 2 more answers
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