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Nadusha1986 [10]
3 years ago
10

A ball is thrown at 25 m/s at an angle of 55 degrees.

Physics
1 answer:
yuradex [85]3 years ago
4 0

Answer:65 seconds

Explanation:just had the quiz

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A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat
umka21 [38]

Answer:

Av_2 =24\ V/V

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width  = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

                           Av₁ Bw₁ = Av₂ Bw₂

                           96 x 8 = Av₂ x 32

                           Av_2= \dfrac{96\times 8}{32}

                           Av_2 =24\ V/V

the highest gain available under these conditions Av_2 =24\ V/V

6 0
3 years ago
A=i+j-k<br>b=j+k<br><img src="https://tex.z-dn.net/?f=a%20%3D%20i%20%2B%20j%20-%20k%20%20%20%5C%5C%20b%20%3D%20j%20%2B%20k" id="
olga nikolaevna [1]

Answer:

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Explanation:

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4 0
3 years ago
Most ionic bonds form when electrons from____.
Margarita [4]
A bond with elements from B.
6 0
3 years ago
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Two wheels are identical but wheel b is spinning with twice the angular speed of wheel
Murljashka [212]
<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>
6 0
3 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
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