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blsea [12.9K]
2 years ago
6

Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

a lifts her end of the table before Ryan does, as a result, the table makes an angle of 150.0° with the horizontal. Find the components of the cups weight that are parallel and perpendicular to the plane of the table
Physics
1 answer:
jarptica [38.1K]2 years ago
6 0
Calculate the weight of the table through the equation,

   W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
   <em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
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A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he
Flura [38]

Answer:

T = 480.2N

Explanation:

In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.

The forces on the boxes are:

T-Mg=0      (1)

T: tension of the rope

M: mass of the boxes 0= 49kg

g: gravitational acceleration = 9.8m/s^2

The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.

By using the equation (1) you obtain:

T=Mg=(49kg)(9.8m/s^2)=480.2N

The woman needs to pull the rope at 480.2N

8 0
2 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
2 years ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
2 years ago
Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th
Inessa [10]

Answer:

F = 2.01*10^-16N -^k

Explanation:

In order to calculate the magnetic force perceived by the bee, you use the following formula:

F=qv\ X\ B            (1)

q: charge of the bee = 1pC = 1*10^-12 C

The average speed of a bee and the magnetic field of the earth are:

v = 6.70m/s

B = 30*10^-6 T

The bee is flying to the west (-^i). You consider that the magnetic field direction is to the north (^j). Then, the direction of the magnetic force is:

-^i X ^j = -^k

You replace the values of the parameters in the equation (1), in order to calculate the magnitude of the force:

|F|=qvB=(1*10^{-12}C)(6.70m/s)(30*10^{-6}T)=2.01*10^{-16}N

The magnetic force perceived by the bee is 2.01*10^-16N in the -^k direction, that is, toward the ground

5 0
2 years ago
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