A soccer ball was kicked . it had the mass of .42 kg and accelerated at 25m/s . what was the force?

A nebula, or swirling cloud of interstellar dust and gases, is seen. What happens next in the star formation process?

AVprozaik [17]

Atoms begin to gravitate together to form a center

5 0

3 years ago

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Which statement about oceans is incorrect?

Hitman42 [59]

It is C because less than one percent of water is fresh water

7 0

2 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

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

or

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

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

2 years ago

An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the sign

Temka [501]

Answer:

Explanation:

Given that,

Frequency of radio signal is

f = 800kHz = 800,000 Hz.

Distance from transmitter

d = 8.5km = 8500m

Electric field amplitude

E = 0.9 V/m

The average energy density can be calculated using

U_E = ½•ϵo•E²

Where ϵo = 8.85 × 10^-12 F/m

Then,

U_E = ½ × 8.85 × 10^-12 × 0.9²

U_E = 3.58 × 10^-12 J/m²

The average electromagnetic energy density is 3.58 × 10^-12 J/m²

6 0

3 years ago

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Mechanical energy is the sum of energy and potential energy.

MAVERICK [17]

Answer:

True, the total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply referred to as the total mechanical energy.

Explanation: Hope it helps you:))))

have a good day

7 0

1 year ago