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2 years ago

The volumes of the block is 10,500 mL and density is 8.52 g/mL. Find the block's mass Don't forget units and a space

Physics

1 answer:

Schach [20]2 years ago

3 0

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 8.52 × 10,500

We have the final answer as

Hope this helps you

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A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet

arlik [135]

The correct answer is 53 meters per second squared

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3 years ago

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The distance between the centers of the wheels of a motorcycle is 146 cm. The center of mass ofthe motorcycle, including the rid

Vitek1552 [10]

Answer:

9.12267515924 m/s²

Explanation:

Here the moment created by the wheels and the moment created by the center of gravity will balance each other.

h = Height of the center of mass = 78.5 cm

d = Distance from back wheel to the center of mass = $\dfrac{146\times 10^{-2}}{2}\ m$

g = Acceleration due to gravity = 9.81 m/s²

a = Horizontal acceleration

The equation is of the form

$mgd=Fh\\\Rightarrow mgd=mah\\\Rightarrow a=\dfrac{gd}{h}\\\Rightarrow a=\dfrac{9.81\times \dfrac{146\times 10^{-2}}{2}}{78.5\times 10^{-2}}\\\Rightarrow a=9.12267515924\ m/s^2$

The horizontal acceleration of the motorcycle that will make the front wheel rise off the ground is 9.12267515924 m/s²

8 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr

svp [43]

The acceleration of the particle at time t is:
$a(t)=24t^2 ft/s^2$
The velocity of the particle at time t is given by the integral of the acceleration a(t):
$v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s$
and the position of the particle at time t is given by the integral of the velocity v(t):
$x(t)=\int v(t) = \int (8t^3)=8 \frac{t^4}{4}=2t^4 ft$

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
$x(2 s)=2 t^4 = 2 (2s)^4=32 ft$

5 0

3 years ago

A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r

sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana $m_{b}$ = 8.97 kg

at distance r = 4/5 { radius of platform }

mass of monkey $m_{m}$ = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + $m_{b}$ ( $\frac{4}{5}$ R)² + $m_{m}$ R² ]w

m/2 × ω₀ = [ m/2 + $m_{b}$ ( $\frac{4}{5}$ )² + $m_{m}$ ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0

3 years ago