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2 years ago

16x^2y^2-25a^2b^2factorize the expression

Physics

1 answer:

SIZIF [17.4K]2 years ago

8 0

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

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A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler

KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the mass of the block:

We replace dta in the formula (1)

F = m*a

100 = m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t Formula (3)

Where:

d:displacement in meters (m)

t : time interval in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0

2 years ago

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f

sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

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2 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

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2 years ago

Which one of these is an exanple of precepation?

Leokris [45]

Answer:

Snow

Explanation:

Precipitation is the formation of a solid after being a liquid. Snow, which is a solid, forms from water, a liquid.

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2 years ago

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What is the relation between the weight of a body and acceleraton due to gravity?

emmainna [20.7K]

Answer:

All objects on Earth, regardless of their mass, accelerate due to gravity at the same rate - that is, 9.8 m/sec2. The weight of an object can be calculated using the formula for force - F = m * a - where F equals the weight of the object and now the acceleration (a) is the acceleration of gravity (g).

6 0

2 years ago

16x^2y^2-25a^2b^2factorize the expression​

16x^2y^2-25a^2b^2factorize the expression