2 years ago

A runner moving at a speed of 8.8m/s rounds a bend with a radius of 25m. what is the centripetal acceleration of the runner

1 answer:

8 0

Radial acceleration is = v^2/r
Therefore (8.8m/s)^2/(25m)= 3.10m/s^2

Hope this helps you! Any questions please just ask! Thank you!

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The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.

dedylja [7]

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

Voltage = 230V

Power = 1500W

To find the resistance, R;

Power = V²/R

Where:

V is the voltage measured in volts.

R is the resistance measured in ohms.

Substituting into the equation, we have;

1500 = 230²/R

Cross-multiplying, we have;

1500R = 52900

R = 52900/1500

R = 35.27 Ohms.

Therefore, the resistance which the heating element needs to have is 35.27 Ohms.

4 0

2 years ago

As the amount of surface area increases, what happens to the rate of this reaction? If the antacid tablet were broken into eight

dangina [55]

as the surface area increases the rate of reaction also increases.

Explanation:

4 0

2 years ago

the frequency of the middle b note on a piano is 493.88 hz. what is the wavelength of this note in centimeters? the speed of sou

vekshin1

Answer:

69.5

Explanation:

V=FX

343.06=493.88×X

X=343.06/493.88

X=0.695m

to cm is 69.5cm

5 0

2 years ago

If the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force

just olya [345]

The object would stay constant

5 0

2 years ago

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

denpristay [2]

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e = 5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y = 11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

$Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa$

Therefore, the tensile stress on the wire is 550 MPa.

8 0

2 years ago