Protons and neutrons grouped in a specific pattern

Maintenance __________ are needed in great numbers to service all sorts of technical equipment.

damaskus [11]

Maintenance spanner are needed in great numbers to service all sorts of technical equipment

5 0

3 years ago

A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien

SSSSS [86.1K]

Answer: $6.86 m/s^2$

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof $\mu _s= 0.7$

Coefficient of kinetic Friction $\mu _k=0.5$

maximum car acceleration is $\mu \times g$

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

$a_{max}=0.7\times 9.8=6.86 m/s^2$

8 0

3 years ago

You drive at 195 miles down the 101 freeway at 65 miles per hour. How long did it take? time =

trasher [3.6K]

Answer:

According to my calculations, if you drive an average of 65 miles per hour, with 0 minutes of stop-time, you should reach your destination in 3 hours and 0 minutes.

Explanation:

7 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago

An ice skater slides in a straight line across the ice. As she passes a certain point, her velocity is 24 km/hr. Two seconds lat

gulaghasi [49]

Acceleration = (change in speed) / (time for the change)

Change in speed = (later speed) - (earlier speed) = (13 - 24) = -11 km/hr
Time for the change = 2 seconds

Acceleration = (-11 km/hr) / (2 sec)

Acceleration = -5.5 km/hr-sec (B)

5 0

3 years ago