A 50 kg child on a skateboard experiences a 75-N force as shown.What is the expected acceleration of the child?

F = 75 N
7
.


A. 0.67 m/s2
B. 1.50 m/s2
C. 6.70 m/s2
D. 25.0 m/s2

Answers

Answer 1
Answer:

Answer:

1.50 m/s²

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a = (f)/(m) \n

f is the force

m is the mass

From the question we have

a = (75)/(50) = (3)/(2) \n

We have the final answer as

1.50 m/s²

Hope this helps you


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Average wavelength of radio waves​

Answers

The average wavelength of radio waves ​ranges from roughly two millimeters to more than 150 kilometers. The wavelengths of radio waves are the longest in the electromagnetic spectrum

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

They also have the lowest frequencies, ranging from around 4,000 cycles per second, or 3 kilohertz, to roughly 280 billion hertz, or 280 gigahertz.

The wavelengths of radio waves are the longest in the electromagnetic spectrum, ranging from roughly two millimeters to more than 150 kilometers.

To learn more about wavelength from here, refer to the link given below;

brainly.com/question/7143261

#SPJ6

Answer:

Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.

Answers

Answer:

-4.40

Explanation:

explanation is in attachment

Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = 3* 10^8\ m/s

b = Wien's displacement constant = 2.897* 10^(-3)\ mK

T = Temperature

From Wien's law we have

\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m

Frequency is given by

\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz

For Halogen

\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m

Frequency is given by

\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz

The maximum frequency is produced by Halogen bulbs which is closest to the value of5.5* 10^(14)\ Hz

Ratio

(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294

The ratio of Incandescent to halogen peak frequency is 0.85294

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}

The momentum (P₂) after a collision will be

\rm P_2 = (m_1 + m_2) u \ cos\  \theta \  \hat{x} \ + (m_1 + m_2) u \ sin \  \theta \  \hat{y}

Applying momentum conservation, we have

\rm  (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\  \theta \   \n\n  ...1

\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \  \theta  ...2

From equations 1 and 2, we have

\rm \theta =  tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta =  tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o

From equation 1, we have

\rm u =    ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\  \theta )  \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s

Then the change in kinetic energy, we have

\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J

The fraction of kinetic energy lost will be

\rm Energy \ lost = (239.34)/(267.5) = 0.895

More about the collision link is given below.

brainly.com/question/13876829

Answer:

Detailed solution is given below

The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery.Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

2.429783984* 10^(-14)\ A

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = 1.6* 10^(-19)\ C

Number of electrons passing per second

n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588

Number of protons passing per second

n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749

Current due to electrons

I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A

Current due to protons

I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A

Total current

I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A

The average current is 2.429783984* 10^(-14)\ A