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ELEN [110]
2 years ago
10

Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.

Physics
1 answer:
Tanya [424]2 years ago
7 0
100cos30° + 80cos120° + 40cos233° + Dx = 0 
<span>Dx = -22.53 N </span>

<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>

<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>

<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
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The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
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Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!
Dominik [7]

the purpose of fuses and circuit breakers is (first answer)

7 0
2 years ago
Read 2 more answers
Determine the frequency of light whose wavelength is 4.257 x10-7 cm
Bas_tet [7]
<span>Frequency x Wavelength = Speed of light Now, speed of light = 3 x 10^5 km/s = 3 x 10^8 m/s = 3 x 10^10 cm/s Frequency = speed/Wavelength = (3 x 10^10)/(4.257 x 10^-7) =7 x 10^16 Hz</span>
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3 years ago
A referee will toss up the ball between to opponents.what is this called
OlgaM077 [116]
This happens in basketball. It is known as "jump ball".
5 0
3 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
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