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3 years ago

Calculate the potential energy of a 4 kg cat crouched 3 meters off the ground

Physics

2 answers:

AnnZ [28]3 years ago

8 0

Answer : Potential energy of the cat is 117.6 J.

Explanation :

It is given that,

Mass of the cat, m = 4 kg

Height from the ground, h = 3 m

We know that the potential energy of the cat is possessed by the virtue of its height under the action of gravity.

Mathematically, it is given as :

$PE=mgh$

$PE=4\ kg\times 9.8\ m/s^2\times 3\ m$

$PE=117.6\ J$

Hence, this is the required solution

Tom [10]3 years ago

7 0

Gravitational potential energy = mass × gravity × height

Ep = (4)(9.81)(3)

Energy = 117.72 Joules

= 1.2x10^2 Joules

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2 years ago

A 2.3 m -long wire carries a current of 8.6 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the

Volgvan

Answer:

Explanation:

Given that :

The current I = 8.6 A

The length vector L = 2.3 m i

The force vector is = -2.0 j N

When L = 2.3 m i ; the

Force vector F = (2.0 i - 4.8 k) N

Compute the components of the magnetic field as follows:

$F = I(L*B)$

Replacing 8.6 A for I ; -2.0 j N for F & 2.3 m i for L

$-2.0 j N = (8.6 A) (2.3 m i *(B_xi + B_yj+B_zk)$

$-2.0 j N = 19.78B_z \ and \ 0 = 19.78B_y$

$B_z = -0.1011 T \ and \ B_y = 0$

However in y direction ; we have :

$(2.0 i - 4.8 k) = 8,6 A (2.3 mi*(B_xi+B_yj+B_zk)$

$- 2.0 = 19.78 B_z \ and \ -4.8 = 19.78 B_x$

$B_z = -0.1011 T \ and \ B_x = -0.2427T$

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5 0

2 years ago

a light bulb has a resistance of 360 . what is the current in the bulb when it has a potential difference of 120 v across it? 0.

xenn [34]

Presume we are looking for the current:

V = IR

120 = I*360

120/360 = I

1/3 = I

I = 1/3 = 0.333..

Current ≈ 0.33 Ampere.

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3 years ago

Read 2 more answers

A metal wire 1.50 m long has a circular cross section of radius 0.32 mm and an end-to-end resistance of 90.0 Ohms. The metal wir

elixir [45]

Answer:

So after stretching new resistance will be 0.1823 ohm

Explanation:

We have given initially length of the wire $l_1=150m$

Radius of the wire $r_1=0.32mm=0.32\times 10^{-3}m$

Resistance of the wire initially $R_1=90ohm$

We know that resistance is equal to $R=\frac{\rho l}{A}$ ,here $\rho$ is resistivity, l is length and A is area

From the relation we can say that $\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}$

Now length of wire become 6.75 m

Volume will be constant

So $A_1l_1=A_2l_2$

So $\pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75$

$r_2=1.508mm$

So $\frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}$

$R_2=0.1823ohm$

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2 years ago

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melisa1 [442]

Answer:

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Explanation:

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2 years ago