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Elina [12.6K]
3 years ago
6

Need help can anyone solve this for me ?

Physics
1 answer:
grin007 [14]3 years ago
4 0
I think its southeast 
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Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?
Minchanka [31]

For each load,  Work = (mass) x (gravity) x (distance .

Bigger load:      Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load:    Work = (5 kg)  x  (9.8 m/s²)  x  (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of  2:1  is exactly balanced by
the height ratio of  1:2 .

6 0
3 years ago
How do you find the change in potential energy
ratelena [41]

P.E = mgh

This is the formula for potential energy.

This is where m is mass, g is the acceleration due to gravity, and h is height.

All you have to do is multiply all these numbers together.

3 0
3 years ago
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is
user100 [1]

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

6 0
4 years ago
Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tig
Andru [333]
I=2.27 x 10 to the 3rd power A
5 0
3 years ago
If f(x)=4/x+2 and g is the inverse of f,then g'(10)=​
ch4aika [34]

Answer:

g'(10) = \frac{-1}{16}

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x    <em> </em><em>(Identity)</em>

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = \frac{1}{f'(x)}    --equation[1]    Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = \frac{4}{x} + 2

\frac{4}{x} = 8

x = \frac{1}{2}

Since f(x) = \frac{4}{x} + 2

f'(x) = -\frac{4}{x^{2} }

f'(\frac{1}{2})  =  -4 × 4 = -16            

Putting it in equation 1, we get:

We get g'(10) = -\frac{1}{16}

5 0
3 years ago
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