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soldier1979 [14.2K]
2 years ago
14

A ball is dropped from a cliff and falls a distance of 20 m to the ground. Determine the velocity it hits the ground at and the

time for the ball to fall.
A.v = -10 m/s, t = 2 s
B.v = 20 m/s, t = 4 s
C.v = -20 m/s, t = 2 s
D.v = 10 m/s, t = 2 s
Physics
1 answer:
Norma-Jean [14]2 years ago
5 0
Use the kinematic equation: d = vi • t + ½ • a • t^<span>2
</span>
<span>20=0(t)+(0.5)(9.8)(t^2)
</span>20=4.9(t^2)
t=2.02seconds
---------------------------------
Final Velocity=Initial Velocity +at^2
Vf=(9.8)(2.02^2)
Vf=19.796
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A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m à 3.00 m à 5.00 m. 1. what is the
cupoosta [38]
1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2:
W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
W_a = W- \rho_W V_{disp} g
where \rho _W = 1000 kg/m^3 is the water density and V_{disp} is the volume of displaced fluid.
The density of the raft (650 kg/m^3) is smaller than the water density (1000 kg/m^3), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
W=B=\rho _W V_{disp} g
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
B=\rho_W V_{disp} g
When the raft is completely immersed in the water, the volume of fluid displaced V_{disp} is equal to the volume of the raft, V_{disp}=V. Therefore the buoyancy in this situation is
B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
B=W=1.91 \cdot 10^5 N

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
mg = m_{disp} g
where m_{disp}= \rho_W V_{disp} is the mass of the displaced water. From the previous equation, we obtain that m_{disp}=m=19500 kg.

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV we can rewrite m=m_{disp} as:
d V =d_W V_{disp}
and so
V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3

8) The volume of water displaced is (point 7) 19.5 m^3. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3 is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3:
\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3. Therefore, the fraction of volume of the raft above water level is
\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%

9) Let's repeat steps 5-8 replacing \rho _W, the water density, with \rho_E=806 kg/m^3, the ethanol density.

9-5) The buoyant force is given by:
B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft: 
m_E = m = 19500 kg

9-7) Using the relationship d= \frac{m}{V}, we can find the volume of displaced ethanol:
V_E =  \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: 24.2 m^3. Therefore, the fraction of raft's volume below the ethanol level is 
\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%
Consequently, the raft's volume above the ethanol level is 
30 m^3 - 24.2 m^3 = 5.8 m^3
and the fraction of volume above the ethanol level is
\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%
8 0
2 years ago
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0
2 years ago
A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
2 years ago
How long does it take for Saturn's equatorial flow, moving at 1500km/h, to encircle the planet?
Ludmilka [50]

Answer:

252.45 hours or 908820 seconds

Explanation:

The equatorial radius of Saturn is 60,268 km

The length of the equator will be circumference

2\pi 60268

Speed of the equatorial flow = 1500 km/h

Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{2\pi 60268}{1500}\\\Rightarrow Time=252.45\ h=252.45\times 60\times 60=908820\ s

It will take 252.45 hours or 908820 seconds for the equatorial flow to encircle the planet.

4 0
1 year ago
Read 2 more answers
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
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