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anygoal [31]
3 years ago
10

A student is holding a marble in his hand. What are the equal and opposite forces that are acting on it?

Physics
1 answer:
Nesterboy [21]3 years ago
4 0

Answer:

Gravity and his hand

Explanation:

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A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The
Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

V_f^2=V_i^2-2gh

By putting the values

0^2=1.9^2-2\times 10\times h

h=0.1805 m

h=18.05 cm

4 0
3 years ago
Which of the following statements are true about the international system of measurement?
anygoal [31]
The International System Units or the SI units is  scientific method of expressing the magnitudes or quantities of important natural phenomena. There are seven base units in the system, from which other units are derived. This system was formerly called the meter-kilogram-second (MKS) system.
8 0
3 years ago
What happens to thermal energy when Kinetic energy is Increased/Decreased?
Elis [28]

Answer:If kinetic energy increases, so does the thermal energy, and vice versa.

Please brainliest!

3 0
3 years ago
A horizontally oriented pipe has a diameter of 5.8 cm and is filled with water. The pipe draws water from a reservoir that is in
Gnoma [55]

Answer:

v₂ = 97.4 m / s

Explanation:

Let's write the Bernoulli equation

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Index 1 is for tank and index 2 for exit

We can calculate the pressure in the tank with the equation

        P = F / A

Where the area of ​​a circle is

       A = π r²

E radius is half the diameter

      r = d / 2

      A = π d² / 4

We replace

    P = F 4 / π d²2

    P₁ = 397 4 /π  0.058²

    P₁ = 1.50 10⁵ Pa

The water velocity in the tank is zero because it is at rest (v1 = 0)

The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa

Since the pipe is horizontal y₁ = y₂

We replace on the first occasion

   P₁ = P₂ + ½ ρ v₂²

  v₂ = √ (P1-P2) 2 / ρ

  v₂ = √ [(1.50-1.013) 10⁵ 2/1000]

  v₂ = 97.4 m / s

6 0
3 years ago
The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

n_{T} \approx 54.842\,rev

5 0
3 years ago
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