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2 years ago

8

Superman is standing 393m horizontally away from Lois Lane. A villian drops a rock 4 m directly above Lois. I f Superman is to i

ntervene and ctath the brock

1 answer:

Nat2105 [25]2 years ago

5 0

You have to draw a triangle

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Which condition must exist in order for conduction to occur between two substances

motikmotik

Heat!!!!!!!!!!!!!!!!!!! sorry about the exclamation marks but it wont let me post in less it has more than 20 characteristics

6 0

2 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

2 years ago

. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to

d1i1m1o1n [39]

Answer:

magnitude of net magnetic field at given point is

$B = 5 \times 10^{-6} T$

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

$B = \frac{\mu_o i}{2\pi r}$

here we we will find the magnetic field due to wire which is along x axis is given as

$i = 30 A$

r = 2 m

now we have

$B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}$

$B_1 = 3\times 10^{-6} T$ into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

$i = 40 A$

r = 2 m

now we have

$B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}$

$B_2 = 4\times 10^{-6} T$ along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

$B = \sqrt{B_1^2 + B_2^2}$

$B = 5 \times 10^{-6} T$

5 0

3 years ago

Is constant velocity same as constant acceleration?

irakobra [83]

Answer:

No

Explanation:

constant velocity means zero acceleration because acceleration is rate of change of velocity. So when velocity remains constant there would be no acceleration.

in case of constant acceleration it means that velocity changes but with some consistent (same) amount each time. For example some object falling from tower, Velocity of the object will change(increases) by 9.8m/s for each second due to gravity of Earth. In this case there is change in velocity each second but with some consistent amount that is 9.8m/s.

7 0

1 year ago

Read 2 more answers

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

2 years ago