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3 years ago

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Physics

1 answer:

Naya [18.7K]3 years ago

4 0

Answer:

0.242

Explanation:

m = 13.3 kg

a =d= 2.42 m/s²

g = 10 m/s²

from the laws of friction F = ¶R

===> ¶ = F/R = ma/mg = a/g

¶ = a/g = 2.42/10 = 0.242

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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

d1i1m1o1n [39]

1) A) Object 1 has the greater momentum

The magnitude of the momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its momentum is

$p_1 = m_1 v_1 = (2m)(v)=2mv$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its momentum is

$p_2 = m_2 v_2 = (m)(\sqrt{2} v)=\sqrt{2}mv$

So we see that $p_1 > p_2$ , so object 1 has the greater momentum.

2) The objects have the same kinetic energy.

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its kinetic energy is

$K_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2m)(v)^2=mv^2$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its kinetic energy is

$K_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(m)(\sqrt{2} v)^2=mv^2$

So we see that $K_1 =K_2$ , so the objects have same kinetic energy

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3 years ago

Three metal spoons are on a table. They are each at room temperature. If the three spoons touch.

hoa [83]

If the three spoon touch nothing happens because they are all at room Temperature

3 0

3 years ago

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A cube has sides that are each equal to 7 centimeters in length. What is the volume of the cube?

RUDIKE [14]

Answer:

Explanation:

7³ = 343

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3 years ago

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The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m

r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0

3 years ago