A weightlifter does 450 J of work on a barbell in 3 s. How much power is the weightlifter generating?

What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?

rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

$E_{k}=\frac{1}{2} *m*v^{2}$

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

For the person running

$E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]$

For the bullet

$E_{k} =\frac{1}{2} *m*v^{2}$

$E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]$

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0

2 years ago

Alondra applies 24 N of force to the end of a stick (a type of lever) to open a can of paint.

Assoli18 [71]

Answer:

1/4

Explanation:

Mechanical Advantage = Load/Effort

Given

Effort applied = 24N

Load = 6N

Substitute

MA = 6/24

MA = 1/4

Hence the mechanical advantage is 1/4

8 0

2 years ago

Read 2 more answers

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

Paul [167]

Answer: $1.23\ m/s^2$

Explanation:

Given

At an elevation of $y=34.7\ km$ , spacecraft is dropping vertically at a speed of $u=293\ m/s$

Final velocity of the spacecraft is $v=0$

using equation of motion i.e. $v^2-u^2=2as$

Insert the values

$\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2$

Therefore, magnitude of acceleration is $1.23\ m/s^2$ .

8 0

2 years ago

Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.

juin [17]

There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵

3 0

3 years ago

A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force

SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, $\theta=30^{\circ}$

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

$F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N$

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0

3 years ago