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Nataly [62]
3 years ago
8

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

er normally flows at a speed of 1 meter per second. What must its speed be in order to transport the particles that are twice as massive as usual?
Physics
1 answer:
kow [346]3 years ago
5 0

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

1^6 = 1 kg

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

2^{\frac{1}{6}} = 1.122 m/s

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Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a
Ann [662]

Answer:

The magnitude of displacement is 0.082m

Explanation:

While the ball is in motion,we have MV + mv= 0 ...eq1

Where M = combined mass of the platform and the two people.

V = velocity of the platform

m = mass of the ball

v = velocity of the ball

The distance that the platform moves is given by:

X = Vt ...eq2

Where t is the time that the ball is in the air.

The time the ball is in the air is given by:

L/(v-V) ...eq3

Where L is the length of the platform

The quantity(v-V) = velocity of the ball relative to the platform.

Combining eq2 and eq3

X = (V/(v - V))L

From eq1 , the ratios of the velocities is V/v = -m/M

X = (V/v)L / (1 - (V/v) = (-m/M) L /(1+ (m/M))

X = -mL/(M + m)

X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)

X = - 10.48/ 122.36

X = -0.082m

The minus sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.

7 0
3 years ago
Read 2 more answers
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
How to use kinetic energy to calculate velocity
BartSMP [9]

As we know the formula of kinetic energy is

KE = \frac{1}{2} mv^2

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

150,000 = \frac{1}{2} * 120 * v^2

v^2 = 2500

v = 50 m/s

So speed of above object is 50 m/s

7 0
3 years ago
What is the equation for momentum? Please help need answer ASAP.
Leona [35]
Momentum is mass in motion and only applies to objects in motion. It's a term that describes a relationship between the mass and velocity of an object, and we can see this when it is written in equation form, p = mv, where p is momentum, m is mass in kg and v is velocity in m/s.
3 0
3 years ago
Read 2 more answers
True or false the potential energy of a freely object increases as it begins to fall
Eva8 [605]
False, as an object falls its potential energy turns into kinetic energy thus decreasing the potential energy.
3 0
3 years ago
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