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koban [17]
3 years ago
13

How Do I get A Picture For My Profile, It keeps saying my pistures won't work

Physics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

So whenever it won't work, try to find a new pic because your current one is too small / big. Type in the size, for example, 10x10 cool backgrounds, and pick one. You can do any size and any background so long as it fits. Also don't multiply 10x10 whenever you're dealing with backgrounds.

Explanation:

I hope this helped!

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Health professionals can help those with health issues to __________.
erastovalidia [21]

Answer:

B

Explanation:

8 0
3 years ago
Read 2 more answers
HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​
sleet_krkn [62]

Answer:

a. 25000 J

b. 2500 J/s

Explanation:

Given,

Distance ( s ) = 50 m

Force ( f ) = 500 N

a.

To find : -

Work done ( W ) = ?

Formula : -

W = fs

W

= 500 x 50

= 25000 J

Therefore,

the work done by the force the horse exerts is

25000 J.

b.

To find : -

Power ( P ) = ?

Formula : -

W = Pt

P = W / t

P

= 25000 / 10

= 2500 J/s

Therefore,

the power produced if the movement took 10 s

is 2500 J/s.

3 0
3 years ago
Read 2 more answers
What happens to light when it strikes the inside surface of a smooth, curved mirror?
grandymaker [24]
<span>It bounces back toward a single spot.  This is how the focusing mirrors on headlights work</span>
3 0
3 years ago
Read 2 more answers
A spring with a spring constant of 165 N/m is attached to a 2.0 kg mass and set into motion.
marin [14]

Explanation:

We have,

Spring constant of the spring, k = 165 N/m

Mass, m = 2 kg

It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

T=2\pi \sqrt{\dfrac{m}{k} }\\\\T=2\pi \sqrt{\dfrac{2}{165} }\\\\T=0.69\ s

The frequency of vibration is reciprocal of its time period. So,

f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.69}\\\\f=1.44\ Hz

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.

3 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
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