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dimulka [17.4K]

2 years ago

7

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

.97 x 10 24kg, radius of the earth = 6.37 x 10 6m)

1 answer:

snow_lady [41]2 years ago

4 0

Answer:

9.81N

Explanation:

the force of attraction is given by

F=<u>GmM</u><u>/</u><u>R²</u><u> </u>

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

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The temperature of a body of water influences _____

Maslowich

The temperature of the air above it

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2 years ago

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(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

son4ous [18]

Answer:

12.5752053801 m/s

$80\times 10^{-3}\ m^3/s$

No.

Explanation:

Q = Volume flow rate = $80\ L/s=80\times 10^{-3}\ m^3/s$

d = Diameter of pipe = 9 cm

A = Area = $\dfrac{\pi}{4}d^2$

Volume flow rate is given by

$Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s$

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is $80\times 10^{-3}\ m^3/s$

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0

3 years ago

To cross the river a swimmer chooses the direction minimizing the amount of time spent in the water. The swimmer swims at a cons

Lynna [10]

Answer:

304 meters downstream

Explanation:

The given parameters are;

The speed of the swimmer = 2.00 m/s

The width of the river = 73.0 m

The speed of the river = 8.00 m/s

Therefore;

The direction of the swimmer's resultant velocity = tan⁻¹(8/2) ≈ 75.96° downstream

The distance downstream the swimmer will reach the opposite shore = 4 × 73 = 304 m downstream

The distance downstream the swimmer will reach the opposite shore = 304 m downstream

6 0

3 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

5 0

2 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago