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Vitek1552 [10]
3 years ago
6

Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo

rce one as a vector in the direction of the positive x-axis and force two as a vector at an angle θ with the positive x-axis.)
Physics
1 answer:
dedylja [7]3 years ago
7 0

Answer:

\theta = 76.9 degree

Explanation:

As we know that the resultant of two vectors is given as

R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta

here we know that

R = 87 Lb

F_1 = 41 Lb

F_2 = 68 Lb

now we have

87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta

87^2 = 6305 + 5576 cos\theta

\theta = cos^{-1}(\frac{1264}{5576})

\theta = 76.9 degree

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A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected
Ganezh [65]

Answer:

The ball's initial kinetic energy

The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy

Explanation:

A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.

From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.

Ki+Ui=Kf+Uf

Ki=initial kinetic energy

Ui=initial potential energy

Kf=final kinetic energy

Uf=final potential energy

we know that \frac{1}{2} mu^{2} +mgha=\frac{1}{2} mv^{2} +mghb

m=mass of the ball

ha=downward height a

hb=upward height b

u=initial velocity u

v=final velocity v, which is 0

g=acceleration due to gravity

v=0 at final velocity

1/2mu^2+mgha=0+1/2mv^2

ha=hb+Ki/mh

From the above equation, we can conclude that the ball's initial kinetic energy  is responsible for making the ball reach point B.

Point B is higher than point A from the motion gained by the ball

3 0
3 years ago
How do you relate and explain a planet's distance from the Sun to things like the time of their year & their average tempare
anzhelika [568]
What we call a "year" is the time a body takes to complete one orbital revolution
in its path around the sun.  The way gravity works, the farther a planet is from the
sun, the slower it moves, and the longer it takes to complete that trip.  So, farther
out from the sun means a longer "year".

Everybody knows that if you want to get more warmth, then you have to stand closer
to the fire, and it's the same with planets.  The farther a planet is from the sun, the less
heat it gets from the sun, and in most cases, that means its average temperature is
lower. (The planet's average temperature is affected by other things besides its distance
from the sun, such as how much heat comes up from inside, and how much heat its
atmosphere traps.)

The farther a planet's rotation axis is tilted from being perpendicular to the plane
of its orbit, the more seasonal variation there can be in the temperature at any one
place on its surface.  Of course, this is kind of irrelevant if the planet has no surface.
5 0
3 years ago
How are credit unions different from banks?
aleksandrvk [35]
Answer: B ≈ Credit unions are owned by stockholders rather than partners
7 0
3 years ago
Linear Thermal Expansion (in one dimension)
Leokris [45]

Answer:

1) \Delta L= 0.612\ m

2) a. \Delta V_G=0.57\ L

   b. \Delta V_S=0.021\ L

   c. V_0=0.549\ L

Explanation:

1)

  • given initial length, L=1275\ m
  • initial temperature, T_i=-15^{\circ}C
  • final temperature, T_f=25^{\circ}C
  • coefficient of linear expansion, \alpha=12\times 10^{-6}\ ^{\circ}C^{-1}

<u>∴Change in temperature:</u>

\Delta T=T_f-T_i

\Delta T=25-(-15)

  • \Delta T=40^{\circ}C

We have the equation for change in length as:

\Delta L= L.\alpha. \Delta T

\Delta L= 1275\times 12\times 10^{-6}\times 40

\Delta L= 0.612\ m

2)

Given relation:

\Delta V=V.\beta.\Delta T

where:

\Delta V= change in volume

V= initial volume

\Delta T=change in temperature

  • initial volume of tank, V_{Si}=60\ L
  • initial volume of gasoline, V_{Gi}=60\ L
  • initial temperature of steel tank, T_{Si}=15^{\circ}C
  • initial temperature of gasoline, T_{Gi}=15^{\circ}C
  • coefficients of volumetric expansion for gasoline, \beta_G=950\times 10^{-6}\ ^{\circ}C
  • coefficients of volumetric expansion for gasoline, \beta_S=35\times 10^{-6}\ ^{\circ}C

a)

final temperature of gasoline, T_{Gf}=25^{\circ}C

∴Change in temperature of gasoline,

\Delta T_G=T_{Gf}-T_{Gi}

\Delta T_G=25-15

\Delta T_G=10^{\circ}C

Now,

\Delta V_G= V_G.\beta_G.\Delta T_G

\Delta V_G=60\times 950\times 10^{-6}\times 10

\Delta V_G=0.57\ L

b)

final temperature of tank, T_{Sf}=25^{\circ}C

∴Change in temperature of tank,

\Delta T_S=T_{Sf}-T_{Si}

\Delta T_S=25-15

\Delta T_S=10^{\circ}C

Now,

\Delta V_S= V_S.\beta_S.\Delta T_S

\Delta V_S=60\times 35\times 10^{-6}\times 10

\Delta V_S=0.021\ L

c)

Quantity of gasoline spilled after the given temperature change:

V_0=\Delta V_G-\Delta V_S

V_0=0.57-0.021

V_0=0.549\ L

8 0
3 years ago
You throw a softball (of mass 350 g) straight up into the air. it reaches a maximum altitude of 13.5 m and then returns to you.
Karolina [17]
GPE = mgh Where m is the mass of the object (kg), g is acceleration due to gravity (which I will assume to be 9.81ms-2), and h is the height of the object above ground level. This is used in cases where the object is close to the earth, since any change in gravitational force is negligible.

Substituting in the numbers:
GPE=0.35*9.81*13.5=46.35 \\ \boxed{=46.4J \ \ to \ 3sf}
7 0
3 years ago
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