All categories

4 years ago

Please help.

Physics

2 answers:

LenKa [72]4 years ago

7 0

Mu wildest guess is c 1500

maw [93]4 years ago

4 0

Answer:

B. 5000

Explanation:

The tow rope should be strong enough so that it can overcome the frictional force and pull pallet without breaking. In question following data is given:

Weight of pallet = W = 6000lbs

Coefficient of friction = uK = 0.42

The tow rope required to pull pallet = (Weight of pallet) (Coefficient of friction)

= (W)( uK)

= (6000)(0.42)

= 2520 lbs

5000lbs is higher value than 2520lbs. Hence, if we will use 5000lbs tow rope it can easily pull the pallet.

You might be interested in

What is the order of magnitude of the gravitational force between two 1.0 kilogram charges that are positioned 1.0 meter apart?

stiks02 [169]

The answer is -11

Enjoy!

6 0

3 years ago

You do work on something when you lift it against gravity. How does this work relate to gravitational potential energy? If the l

AleksandrR [38]

<span>Lifting an object increases the gravitational potential energy of the system. If you release the object, that potential energy will be transformed into the energy of an object in motion which is termed as the kinetic energy as it falls toward earth. Hope this helps.</span>

6 0

3 years ago

Read 2 more answers

A 15.0 Kg object is moved from a height of 7.00 m above thefloor to a height of 13.0 m above the floor. What is the change ingra

Pie

To solve this problem we will apply the concepts related to potential gravitational energy. This is defined as the product between mass, acceleration and change in height and can be expressed as,

$\Delta PE = mg \Delta h$

Here,

m = Mass

g = Gravitational acceleration

$\Delta h$ = Height

Replacing with our values we have,

$\Delta PE = (15kg)(9.81m/s^2)(13m-7m)$

$\Delta PE = 882.9J \approx 883J$

Therefore the change in gravitational potential energy is 883J.

4 0

3 years ago

Water from a leaking tap begins to fall into

Amanda [17]

Answer:

It will take 20,000 seconds to get a pressure of 200Pa at the bottom of the tank.

Explanation:

The pressure at the bottom of the tank will be

$P = \dfrac{Mg}{A}$

where $M$ is the mass of the water, and $A$ is the base area of the tank.

The base area of the tank is

$A = 0.5m* 2m = 1m^2$ ,

and if we want the pressure at the bottom to be 200pa, then it must be that

$200Pa = \dfrac{M(10m/s^2)}{1m^2}$ ,

solving for $M$ we get:

$M = 20kg\\$

which is the required mass of the water in the tank.

Now, the tank fills at a rate of 2 drops per second or

$2 (0.05g)/s = 0.10g/s = 0.0001kg/s$

since each drop weights 0.05g.

Therefore, the time $t$ it takes to collect 20kg of water will be

$t = 20kg \div \dfrac{0.0001kg}{s}$

$t = 2*10^5s$

which is 55.56 hours or 2 days and 7.56 hours.

Thus, it will take 20,000 seconds to get a pressure of 200Pa at the bottom of the tank.

7 0

3 years ago

In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal.

MariettaO [177]

Answer:

18.7 m

Explanation:

$v_{o}$ = initial speed of the shot = 13 m/s

θ = angle of launch from the horizontal = 47 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 1.80 m

t = time of travel

Using the kinematics equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 1.80 = (9.5) t + (0.5) (- 9.8) t²

t = 2.11 s

Consider the motion along the horizontal direction

x = horizontal displacement of the shot

$v_{ox}$ = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.11 s

Using the kinematics equation

$x=v_{ox} t+(0.5)a_{x} t^{2}$

x = (8.87) (2.11) + (0.5) (0) (2.11)²

x = 18.7 m

8 0

3 years ago