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3 years ago

What direction does an applied force move an object

2 answers:

7 0

Lets say you push a ball from the back, is it gonna come back towards you or go away from you. So an object will go the same direction that the applied for pushed it

Hope this helps :)

monitta3 years ago

4 0

The object will move in the direction of the applied force.

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Which of these are a covalent compound?

nikdorinn [45]

C is a non-metal and so is O. So the answer is CO

8 0

3 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

$D=6.75\times 10^5\ m$

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

$D=1.12\times 10^6\ m$

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

$D=1.68\times 10^6\ m$

Hence, this is the required solution.

7 0

3 years ago

What is the disorder caused when cells lose the ability to control growth and continue to divide?

8090 [49]

The answer is cancer that is the reason there is no complete cue in people

7 0

3 years ago

During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.

swat32

Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

3 0

3 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

Read 2 more answers