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Aleks04 [339]
3 years ago
14

Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he

had 540 dimes and nickels altogether in the end, how many more nickels than dimes did he have ?​
Physics
2 answers:
TiliK225 [7]3 years ago
8 0

Answer:

i think 53 or 56

Explanation:

cause i cross multiplied and got that

FromTheMoon [43]3 years ago
8 0

Answer :

He had 300 more nickels than dimes in the end

Explanation:

The ratio of dimes and nickels has changed to 6:21; we have known this son that is 540 dimes and nickels together the. We can make a proportion difference/sum=15/27= difference/540. By cross multiplying, difference=15 x 540  divided by  27= 300

Which gives us the answer that he had 300 mor nickels than dimes in the end.

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A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
3 years ago
To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

W = 11.5* \frac{\pi}{3}

W = 12.09J

4 0
3 years ago
An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image
Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

(b) The image formed is diminished and real.

<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

Learn more about concave mirror here: brainly.com/question/13164847

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5 0
1 year ago
A car start from rest and move 6km due east and 8km due south. Calculate the distance covered
Kruka [31]

for this you use the pythagoreom theorem

6^2 + 8^2

36 + 64 = 100

the square root of 100 is 10

10 is the answer


5 0
3 years ago
Read 2 more answers
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

3 0
3 years ago
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