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3 years ago

10

A draft is made of a plastic block with a density of 650kg/m^3, and its dimensions are 2.0mx3.0mx5.0m ,what is the raft's appare

nt weight in water?(hint:density of water=1.00x10^3kg/m^3

1 answer:

mixas84 [53]3 years ago

8 0

Weight in water = mass of block - mass of volume of displaced water

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Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the

Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = ( $\frac{1}{5}$ × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂ = 14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0

4 years ago

Summarize renewable energy saves water and creates jobs

siniylev [52]

Answer: Solar Energy has been helping the supplying companies to be able to hire more people for jobs which helps 12 times more easier than the US, Things like solar, geothermal, and wind can help reduce the water resources which helps to save our water conservation rate which was getting increasingly High.

<em>I hope this explanation had helped you today!</em>

4 0

3 years ago

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

3 years ago

Physics Kinematics question

just olya [345]

An interesting problem, and thanks to the precise heading you put for the question.

We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.

Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m

Assume g = -9.81 m/s^2

initial velocity, v m/s (to be determined)

Solution:

(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.

substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s

(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s

Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m

(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m

(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)

vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)

4 0

4 years ago

A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w

Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force = force x displacement

f cos28.3 x d

μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37 J

c )

Work done by frictional force

= frictional force x displacement

= - μ x (mg - f sin28.3 ) x d

= - μ x mgd + f μsin28.3 x d

= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0

3 years ago