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Rashid [163]
2 years ago
10

A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula:

PE = mgh)
3.9 m
4.0 m
39.2 m
40.0 m
Physics
2 answers:
Flura [38]2 years ago
7 0

Answer:

B) 4.0

Explanation:

Jus took the test on E D G E

Tpy6a [65]2 years ago
6 0

Answer:

The answer is 4.0 m

Explanation:

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A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight
LuckyWell [14K]

Answer:

At an angle of 30^{\circ}

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}

6 0
2 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

4 0
3 years ago
A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a
olganol [36]
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
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6 0
3 years ago
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha
leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

6 0
3 years ago
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According to my research I found questions with the same info with option choices and they are:

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Among those options the answer would be the second choice B)
8 0
3 years ago
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