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tankabanditka [31]
3 years ago
7

Imagine that a bowling ball needs to be lifted 1.5 meters, and its gravitational potential energy is 90 joules. How much does th

e bowling ball weigh?
Physics
1 answer:
Vikki [24]3 years ago
8 0

Answer: 6.12 kg

Explanation:

Since Mass of ball = ? (let the unknown value be Z)

Acceleration due to gravity, g= 9.8m/s^2

Height, h = 1.5 metres

Gravitational potential energy GPE = 90J

Gravitational potential energy depends on the weight of the ball, the action of gravity and height.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

90J = Z x 9.8m/s^2 x 1.5m

90 = Z x 14.7

Z = 90/14.7

Z = 6.12 kg

Thus, the bowling ball weigh 6.12 kilograms

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If the total mass is m, find the moment of inertia about an axis through the center and perpendicular to the plane of the square
My name is Ann [436]
Question: A thin, uniform rod is bent into a square<span> of... A thin, uniform rod is bent into a </span>square<span> of side length a. </span>If the total mass is M<span>, </span>find the moment of inertia about an axis through the center and perpendicular to the plane of the square<span>. </span>Use the parallel-axis theorem<span>.</span>
5 0
3 years ago
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

3 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
How do you calculate resistance with coulombs, time and J's?
Mars2501 [29]
Hey user


The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):

E(J) = V(V) ×<span> Q</span>(C)

So

joule = volt × coulomb

or

J = V × C

Example

What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?

E = 15V × 4C = 60J

4 0
3 years ago
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