Help please!!

Is it possible for a car to be accelerating to the west while it is driving to the east?

Ber [7]

Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

8 0

3 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

vovikov84 [41]

Gravitational force is given by, $F= G\frac{mM}{R^{2}}$

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, $F_{1}= G\frac{m_{1}M}{R^{2}}$

Gravitational force of the star on planet 2, $F_{2}= G\frac{3m_{1}M}{(3R)^{2}}$

Ratio, $\frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}$

$\frac{F_{1}}{F_{2}}= \frac{3}{1}$

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0

3 years ago

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An 85-kg refrigerator is located on the 70th floor of a skyscraper (4km above the ground ) what is the potential energy

Airida [17]

Potential energy can be calculated using the following rule:
potential energy = mgh where:
m is the mass = 85 kg
g is the acceleration due to gravity = 9.8 m/sec^2
h is the height = 4 km = 4000 meters

Substitute in the above equation to get the potential energy as follows:
Potential energy = 85*9.8*4000 = 3332000 joules

7 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$