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3 years ago

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What is the weight of a ring tailed lemur that has a mass of 10 kg? Use w = mg (g = -9.8 m/s^2)

1 answer:

marusya05 [52]3 years ago

4 0

m = 10 kg

g = -9.8 m/s2

w = m * g

w= -9.8 * 10

w= -98 N

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A magnet falls through a loop of wire with the south pole entering first. After it has fallen all the way through the wire loop

Alisiya [41]

Answer:

The induced current direction as viewed is clockwise

Explanation:

Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current

Therefore, as the magnet approaches the coil with the south pole, the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

Therefore;

The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

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3 years ago

A 15 kg runaway grocery cart runs into a spring with spring constant 230 N/m and compresses it by 56 cm .What was the speed of t

liubo4ka [24]

To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

$KE = \frac{1}{2} mv^2$

And the potential energy is

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

Here,

m = mass

v = Velocity

x = Displacement

k = Spring constant

There is equilibrium, then,

KE = PE

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

Our values are given as,

$x=0.56m\\k=230N/m\\m=15kg$

Replacing we have that

$\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2$

$v = 2.19m/s$

Therefore the speed of the cart is 2.19m/s

3 0

3 years ago

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as

Leona [35]

(a) -48.0 cm, diverging

We can use the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}$

$f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm$

The lens is diverging, since the focal length is negative.

(b) 6.38 mm, erect

We can use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the size of the image

y = 8.50 mm is the size of the object

Substituting p and q that we used in the previous part of the problem, we find y':

$y'=-y\frac{q}{p}=-(8.50 mm)\frac{-12.0 cm}{16.0 cm}=6.38 mm$

and the image is erect, since the sign is positive.

(c)

See attached picture.

4 0

2 years ago

Please answer it.....

aleksklad [387]

There is no question to answer bud.

3 0

3 years ago

A clarinetist, setting out for a performance, grabs his 3.230 kg3.230 kg clarinet case (including the clarinet) from the top of

SpyIntel [72]

Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

8 0

3 years ago