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IrinaK [193]
3 years ago
12

Please do ittt I really need it ​

Physics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

ok bro

Explanation:

jdkdkfnfnfkfn fkfkkfodow icjdbe

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PLS HELP!! YOU CAN SKIP THE INFO IF YOU WANT!!
Oduvanchick [21]

Answer:

B and C

Explanation:

Both of them have to do with the ocean which was mentioned

6 0
3 years ago
Read 2 more answers
A person running has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?
allsm [11]

Explanation:

Momentum is mass times velocity.

p = mv

720 kg m/s = m (5 m/s)

m = 144 kg

7 0
3 years ago
One of the great upcoming sports in the Olympics is the sport of curling. Write a brief essay on the uses of momentum collisions
Svetllana [295]

The Olympic sport of curling is one that is practically designed to show Physics in motion. Curling is a sport in which two teams alternate sliding smoothed stone pucks down an ice rink court with the intent to seat their stone closest to the center of the target (called the house). Each team has eight stones, meaning that the team that goes second has the (could be) massive advantage of sending the last stone.  

The mass of the stone is important in that the more massive a stone (m) and the speed at which it travels (v) dictates it's momentum (momentum=mxv). As the curling stone slides down the ice (which is relatively frictionless unless acted upon by other players or objects) and having inertia, continues in it's straight course (again, unless acted upon by outside forces). If the stone hits another stone, it transfers some of its momentum in an elastic collision to that stone and the original stone is deflected in a calculable manner.    

Collisions are used in the game to either clear opponent's stones from the house or out of their defensive positions, or to make adjustments to one's stones present in the house, all based on the momentum of the moving stone, and its transference.

6 0
3 years ago
In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- ve
choli [55]

Answer:

B_{int}=-0.015T

Explanation:

From the question we are told that:

RF source oscillation speed \sigma= 34 MHz

The external field Bext =0.78 T.

Pro- ton magnetic moment component \mu=1.41 X 10-26 J/T

Generally the equation for magnitude of B_{int} is mathematically given by

B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}

B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}

B_{int}=0.78-0.7957

B_{int}=-0.015T

6 0
3 years ago
A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after
Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=9.81ms^{-2}

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

v=u+at

Using the formula and substitute the values

0=24.1-9.81t

Where g is negative because motion of ball is against gravity

24.1=9.81t

t=\frac{24.1}{9.81}=2.46s

3.y=y_0+ut+\frac{1}{2}at^2

Using the formula

y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2

y=30.1 m

4.Time of flight for red ball=3.77-2.9=0.87s

y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2

y'=22.0m

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2

0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2

0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2

0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t

-2.86105=-2.851t

t=\frac{2.86105}{2.851}=1.004 s

Hence,  1.004 s  after the blue ball is thrown  are the two balls in the air at the same height.

8 0
3 years ago
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