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3 years ago

What is the momentum p if the mass m is doubled? Vzp 1) р 2) 2p 3) 4p

Physics

1 answer:

aliina [53]3 years ago

3 0

Answer:

<em>Correct answer: 2) 2p</em>

Explanation:

<u>Momentum</u>

Momentum is a measurement of mass in motion or how much mass is in how much motion. The symbol for momentum is p and is defined as:

p = m . v

Where m is the mass of the object and v is the speed. Note this is the scalar version of the formula because both p and v are vectors.

Suppose the mass of the object is doubled to 2m, then the new momentum is:

p' = (2m) . v = 2 m . v

Or, equivalently:

p' = 2p

Correct answer: 2) 2p

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A vector has a component of 10 m in the x-direction, a component of 10 m in the y-direction, and a component of 5 m in the z dir

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The magnitude of this vector is 15

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

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vector a = 10 i + 10 j + 5 z

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

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3 years ago