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2 years ago

I am no out of points for i have been giving them all away

Physics

2 answers:

Blababa [14]2 years ago

7 0

Answer: ok

Explanation:

Vsevolod [243]2 years ago

3 0

Thank you kind person

*hands you a whole gallon of ketchup :)))

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3. You are flying 2586 miles from San Francisco to New York. An hour into the flight, you are 600

Citrus2011 [14]

Answer:

268.2 m/s (1dp)

Explanation:

600 miles = 965,606m (1mile = 1609.34m)

965606/60 = 16,0943.43 ..... (metres traveled every minute)

16,0943.43/60 = 268.2238 .....

3 0

3 years ago

Read 2 more answers

The wind pushes a paper cub along the sand at a beach. The cup has a mass of 25 grams (= ? Kg's) and accelerates at a rate of 5

bixtya [17]

Force = (mass) x (acceleration)

= (0.025 kg) x (5 m/s²)

= 0.125 Newton

7 0

3 years ago

A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare

zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

$F = \frac{\Delta P}{\Delta t}$

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

$F_{12} = F_{21}$

so we know that

$\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}$

so we have

$\Delta P_{12} = \Delta P_{21}$

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin

Nat2105 [25]

Answer:

b) Gravity

Explanation:

Gravity acts all of the time, when you apply force to a projectile it has to be more than the forces of the gravity and air resistance together so the projectile can move, when the rock is at the top of its trajectory the force that you applied at the beginning is getting lost, so the other forces (air resistance and gravity) make the rock fall to the floor.

8 0

3 years ago