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Taya2010 [7]
2 years ago
10

The gravitational potential energy of an object is due to

Physics
1 answer:
ExtremeBDS [4]2 years ago
8 0

d. all of these

look at the picture to know from where we get this law with application on it

You might be interested in
Which are causes of mechanical weathering? (check all that apply)
solmaris [256]

The correct answers are:

B. plant growth;

C. animal actions;

The mechanical weathering is a type of weathering where physical force is included into the breaking up of the rocks. The plants and the animals are both causing this type of weathering with their actions. The plants can cause mechanical weathering with their roots, as they grow and surround a rock, they are able to create such a pressure that they can break the rock apart. Also, as their trunks are getting bigger, if there's rocks right next to them, the pressure from the growing of the trunk will crack the rocks. The animals are able to move the rocks, as well as pushing them, or even deliberately throwing them, so they manage to break up parts of them and cause mechanical weathering.

8 0
3 years ago
Read 2 more answers
A satellite orbits at a distance from the Earth's center of about 2.60 Earth radii and takes 5.89 hours to go around once. What
Elenna [48]

Answer:

424088766.068 m

Explanation:

Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R

R = Radius of earth = 6371000 m (mean radius)

In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit

Circumference of a circle = 2×π×r

⇒Distance travelled in 5.89 hours = 2×π×2.6 R

⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000

⇒Distance travelled in 5.89 hours = 104078451.3393m

Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m

∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m

5 0
3 years ago
You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli
andrew11 [14]

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction  

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

 The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction      

3 0
3 years ago
A train is moving with a velocity of 70 km/h. If north is taken as the positive direction, what is the direction of movement of
Anna11 [10]

Answer:

+70 km/h north

Explanation:

4 0
2 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
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