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3 years ago

What is the potential energy of a 10 kg block that is 5 m above the floor?

Physics

1 answer:

siniylev [52]3 years ago

7 0

Answer:

<h3>The answer is 500 J</h3>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 5 × 10 × 10

We have the final answer as

Hope this helps you

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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

4 years ago

A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate

Vadim26 [7]

Answer:

$1.52*10^6 Nm^2/C$

Explanation:

Given that:

Electrical field E = $7.0 * 10^{-4}N/C$

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux $\phi_E$ can now be determined by using the expression

$\phi_E$ = $E*A*Cos \theta$

$\phi_E$ = $7.0 * 10^{-4}N/C$ $*25.0m*Cos 30^0$

$\phi_E$ = $1515544.457 Nm^2/C$

$\phi_E$ = $1.52*10^6 Nm^2/C$

5 0

3 years ago

Read 2 more answers

Count the number of atoms of each element in this molecule: 3He(H20)2 PLEASE HELPP!!!!!!

kykrilka [37]

Answer:

He=3 H=6 O=6

Explanation:

8 0

3 years ago

Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k

boyakko [2]

Answer:

parametric representation: x = u, y = v - u , z = - v

Explanation:

Given vectors :

i - j , j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + <em>u</em>a + vb

where: r₀ = position vector

u and v = real numbers

a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u , z = - v

3 0

3 years ago

A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the

tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0

3 years ago

What is the potential energy of a 10 kg block that is 5 m above the floor?​

What is the potential energy of a 10 kg block that is 5 m above the floor?