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Kazeer [188]
3 years ago
7

Write the following in expanded form. 4.35 x 10^6

Physics
1 answer:
EleoNora [17]3 years ago
5 0
4.35x10x10x10x10x10x10 if you wanted complete expanded form it would be
4+.3+.05x10x10x10x10x10x10
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Which of the following has the most potential energy?
Alex
A Car at the top of a hill.

It is because in that case, produce of mass and height is highest which is directly proportional to potential energy 

In short, Your Answer would be Option A

Hope this helps!
3 0
3 years ago
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Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi
Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0
3 years ago
What do tendons connect skeletal muscles to?​
Doss [256]

Answer:

Tendons connect muscle to bone. These tough, yet flexible, bands of fibrous tissue attach the skeletal muscles to the bones they move. Essentially, tendons enable you to move; think of them as intermediaries between muscles and bones.

Hope this helps! (:

6 0
3 years ago
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The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.
jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

E=\dfrac{1.2}{5.0\times10^{-3}}

E= 240\ N/C

Hence, The electric field value is 240 N/C

6 0
3 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
3 years ago
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