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Marina CMI [18]
2 years ago
15

How much force must be applied for a 150.W motor to keep a package moving at 3.00m/s?

Physics
1 answer:
leva [86]2 years ago
7 0

The force must be applied for motor to keep the package moving is 50 N.

<h3>What is force?</h3>

Force is the action of push or pull the object in order to make it move or stop.

Power is related to force and velocity as

P = F x v

150 W = F x 3 m/s

F  =50 N

Thus, the force must be applied for motor to keep the package moving is 50 N.

Learn more about force.

brainly.com/question/13191643

#SPJ1

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The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t
Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension T_{da} is

T_{DA}=-4.433i-3.49j+2.735k

From the Question we are told that

M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft

\vec {DA}=-4.7i-3.7j+2.9k)ft

\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft

Generally the equation for T_{DA}  is mathematically given as

T_{DA}=\phi_{DA}* M_{da}

Where

\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}

Therefore

T_{DA}=\phi_{DA}* M_{da}

T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27

T_{DA}=-4.433i-3.49j+2.735k

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
Which country or region is coal-rich and oil-poor?
zimovet [89]
THE ANSWER FOR UR QUESTION IS 

WHICH COUNTRY OR REGION IS COAL RICH AND OIL POOR
ANS:- UNITED STATES IS THE COUNTRY WHICH IS COAL RICH AND OIL POOR

HOPE THIS HELPS!!!

^_^ HAPPY TO HELP U ^_^







5 0
3 years ago
The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of
Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV    .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

or

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0
3 years ago
Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
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