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Aneli [31]
3 years ago
9

A baseball thrown from the outfield to home plate does not have which of the following types of energy while it's in the air? A.

Gravitational potential energy B. Mechanical energy C. Kinetic energy D. Radiant energy
Physics
1 answer:
posledela3 years ago
7 0
The ball does not have D. Radiant energy.
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Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate
pychu [463]

Answer:

B)

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

F=W-N=ma

which means

N=W-ma=mg-ma=m(g-a)

and for our values this is:

N=mg-ma=(70kg)(9.8m/s^2-1.7m/s^2)=567N

5 0
3 years ago
Which compound has ionic bonding? A. Cl2 B. HF C. CaO D. NO2
GrogVix [38]

the answer is CaO because that's what my homework says is correct


7 0
3 years ago
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}

E(0.89) = 306500 N/C

3 0
3 years ago
Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
1 year ago
A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm
joja [24]

Answer:17.44A

Explanation: Current density=I/Area

Area is given by 2.79mm^2=2.79×10^-6m^2

Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A

3 0
3 years ago
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