Non living things in the ocean

A jogger runs 5.0 km on a straight trail at an angle of 60 south of west. What is the southern component of the run rounded to t

svet-max [94.6K]

The components are the sides S and W of a right triangle with SW = 5 km as the hypotenuse. It helps to sketch out these types problems.

Using the 60°south of west angle would make W adjacent and S opposite to the angle. using SOH CAH TOA

sin(60) = S/5
5*sin(60) = S
5*√(3)/2 = S
4.3 km = S

7 0

4 years ago

A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. (a) What is their v

nevsk [136]

Answer:

a) 16.86 m/s.

b) 15.40 m/s

c) 3.175 m/s

Explanation:

let pi be the initial momentum of the system, pf be the final momentum of the system, m1 be the mass of the bullet and V1 be the intial velocity of the bullet, m2 be the mass of the block and V2 be the intial velocity of the block

a) from the conservation of linear momentum:

pi = pf

m1×(V1)i + m2×(V2)i = Vf×(m1 + m2)

(0.0220)×(400) + (0.500)×(0) = Vf(0.0220 + 0.500)

8.8 = 0.522×Vf

Vf = 16.86 m/s

Therefore, the bullet-embedded block system will be moving with a speed of 16.86 m/s.

b) let Vf be the final velocity of the bullet-embedded block system, Wf be the work done by friction on the system, Vi be the initial velocity that the bullet-embedded block system initially moves with.

the total work done on the system is given by:

Wtot = Δk = kf - ki

Wf = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

f×ΔxΔcos(Ф) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

m×Δx×g×μ×(-1) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

m×g×Δx×μ = 1/2(Vi)^2 - 1/2(Vf)^2

1/2(0.0220 + 0.500)(Vf)^2 = 1/2(0.0220 + 0.500)(16.86)^2 - (0.0220 + 0.500)×(9.8)×(8)×(0.30)

(0.261)(Vf)^2 = 86.469

(Vf)^2 = 237.2196

Vf = 15.40 m/s

Therefore, bullet-embedded block system will have a speed of 15.40 m/s after sliding of the friction surface.

c) let Vf be the speed of the bullet-embedded before hitting the block, Vi be the initial velocity of the block and V be the velocity of the bullet-embedded block and block system.

M×Vf + m×Vi = (m + M)×V

(0.0200 + 0.50)×(15.40) + (2)×(0) = (0.022 + 0.50 + 2)×V

8.008 = 2.522×V

V = 3.175 m/s

Therefore, the bullet-embedded block and block system will be moving at a speed of 3.175 m/s.

5 0

3 years ago

Gravitational energy can be negative?

AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

6 0

3 years ago

what is the distance moved by a particle moving along x axis if it starts from x=-7m goes upto x= -12m turns and stops at x=+34m

Marizza181 [45]

Answer: Trajectory=51m

Displacement=41m

Explanation:

Let's begin by stating clear that <u>movement is the change of position of a body at a certain time.</u> So, during this movement, the body will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body (is a scalar magnitude).

The displacement is the distance in a straight line between the initial and final position (is a vector magnitude).

According to this, in the description of the object placed at x= -7m on a number line and moving some 12m to the left and then to the right, stopping at x=34m; we are talking about the path followed by the object, hence its <u>trajectory</u>. So, 51 m is its trajectory.

But, if we talk about displacement, we have to draw a straight line between the initial position of the object (x=-7m) to its final position (x=+34m).

Now, being this an unidimensional problem, the displacement vector for this object is 41m.

7 0

3 years ago

Please help me find b) and d):

maks197457 [2]

Answer:

wow it's so hard

8 0

3 years ago