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Lubov Fominskaja [6]

4 years ago

10

A 30%-efficient car engine accelerates the 1300 kg car from rest to 10 m/s . How much energy is transferred to the engine by bur

ning gasoline

1 answer:

WITCHER [35]4 years ago

3 0

Answer:

The Energy transferred to the engine by burning gasoline = <u>216.67 KJ</u>

Explanation:

The parameters given are:

The efficiency of the car engine, E = 30% = 0.3

Mass, m = 1300 kg

Initial velocity, u = 0, since the car is from rest

The final velocity, v = 10 m/s

Since the car was moving, we calculate its kinetic energy.

kinetic energy = ((1/2) (m) (v^2)

((1/2) (1300 kg) (10 m/s^2)

= 65,000 j

The Energy, Q transferred to the engine by burning gasoline in this case

= potential energy / The efficiency of the car engine, E

Q = 65,000 j / 0.3

= 216,666.66 J

Converting Joule to kilojoule

where 1KJ = 1000j

216,666.66 J = <u>216.67 KJ</u>

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The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

DanielleElmas [232]

Answer:

$V_m = 12.78 km/hr$

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

$F_r = \dfrac{V}{\sqrt{L g}}$

for dynamic similarities

$(\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}$

$(\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})$

$(\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})$

$V_m = 12.78 km/hr$

hence, the velocity of model will be 12.78 km/h

6 0

3 years ago

A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes

Lerok [7]

Answer:

R = 715.4 N

L = 166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L = 166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

0 = 0 checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0

3 years ago

5. Is it likely that light with frequencies higher than ultraviolet was

Vinvika [58]

Answer: YES

Explanation:

If Tia Ana exposes her eyes to light ray in which it frequency is higher than ultraviolet ray, it may result to a gradual eyes issue such as cataracts as the radiation damages the cornea and the lens in front of the eyes.

3 0

3 years ago

A loop of wire is placed in a magnetic field such that it has a flux, LaTeX: \phi ϕ, through it. The loop is compressed so that

Juliette [100K]

Answer:

B'= 3.333 B

Explanation:

Lets take

Initial area = A

Magnetic field = B

The area after compression

A'=0.3 A

Magnetic field = B'

We know that flux ,Ф

Ф = B A

Given that flux is constant so

B A = B' A'

B A=B' x 0.3 A

B'= 3.333 B

It means that magnetic field will increase.

6 0

3 years ago

Read 2 more answers

Beginning from rest when = 20°, a 35-kg child slides with negligible friction down the sliding board which is in the shape of a

jolli1 [7]

Answer:

a) $a_{T}=8.50m/s^{2}$ , $v_{T}=2.78 m/s$ and $N=279.83N$

b) $a_{T}=0m/s^{2}$ , $v_{T}=5.68 m/s$ and $N=795.2N$

Explanation:

a)

In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:

$h_{0}=2.5m-h_{1}$

so we get:

$h_{1}=2.5m(sin (20^{o}))$

$h_{1}=0.855m$

so

$h_{0}=2.5m-h_{1}$

$h_{0}=2.5m-0.855m$

$h_{0}=1.645m$

once we got this value, we can find the final height the same way. This time the angle is 30° so we get:

$h_{f}=2.5m-h_{2}$

$h_{f}=2.5m-2.5m(sin 30^{o}))$

$h_{f}=1.25m$

Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:

$U_{0}+K_{0}=U_{f}+K_{f}$

the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:

$U_{0}=U_{f}+K_{f}$

so now we substitute with the corresponding formulas:

$mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}$

if we divided both sides of the equation by the mass, then the equation simplifies to:

$gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}$

and now we can solve for the final velocity so we get:

$v_{f}=\sqrt{2g(h_{0}-h_{f}}$

and we can now substitute values:

$v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}$

which solves to:

$v_{f}=2.78m/s$

which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:

We can build a free body diagram (see attached picture) and do a balance of forces so we get:

$\sum{F_{x}}=ma_{T}$

in this case we only have one x-force which is the x-component of the weight, so we get that:

$w_{x}=ma_{T}$

so we get:

$mgcos\theta=ma_{T}$

and solve for the tangential acceleration so we get:

$a_{T}=gcos\theta$

$a_{T}=9.81m/s^{2}*cos(30^{o})$

$a_{t}=8.50m/s^{2}$

Now we can find the centripetal acceleration by using the formula:

$a_{c}=\frac{V_{T}^{2}}{R}$

$a_{c}=\frac{(2.78m/s)^{2}}{2.5m}$

so we get:

$a_{c}=3.09m/s^{2}$

Next, we can find the normal force which is found by doing a sum of forces on y, so we get:

$\sum{F_{y}}=ma_{c}$

so we get:

$N-w_{y}=ma_{c}$

and we solve for the normal force so we get:

$N=ma_{c}+w_{y}$

and substitute:

$N=ma_{c}+mg sin\theta$

when factoring we get:

$N=m(a_{c}+g sin\theta)$

and we substitute:

$N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})$

which yields:

N=279.83N

b)

The procedure for part b is mostly the same with some differences due to the angle. First:

$h_{0}=1.645m$

$h_{f}=0m$

so

$U_{0}+K_{0}=U_{f}+K_{f}$

in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:

$U_{0}=K_{f}$

so we get:

$mgh_{0}=\frac{1}{2}mv_{f}^{2}$

so we solve for the final velocity so we get:

$v_{f}=\sqrt{2gh_{0}}$

and we substitute:

$v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}$

$v_{f}=5.68m/s$

according to the free body diagram we get that:

$a_{T}=gcos\theta$

$a_{T}=9.81m/s^{2}(cos 90^{o})$

which yields:

$a_{T}=0$

we can also find the centripetal acceleration, so we get:

$a_{c}=\frac{V_{T}^{2}}{R}$

$a_{c}=\frac{(5.68m/s)^{2}}{2.5m}$

so we get:

$a_{c}=12.91m/s^{2}$

and we can do a sum of forces on y to find the normal force:

$\sum{F_{y}}=ma_{c}$

so we get:

$N-w_{y}=ma_{c}$

and we solve for the normal force so we get:

$N=ma_{c}+w_{y}$

and substitute:

$N=ma_{c}+mg$

when factoring we get:

$N=m(a_{c}+g)$

and we substitute:

$N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})$

which yields:

N=795.2N

3 0

3 years ago