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3 years ago

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On average, 10 inches of freshly fallen snow melt down to 1 inch of liquid water. using this 10 to 1 ratio, the water equivalent

depth of 45 cm of fresh snow is about _____ cm.

1 answer:

weeeeeb [17]3 years ago

8 0

Hello

The ratio between depth of snow and depth of water is 10:1. So, if we use 45 cm as depth of snow, the proportion to write is
$10:1=45:x$
where x is the depth of equivalent water. From the proportion we get
$x= \frac{45cm}{10} =4.5$ cm.

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A 62-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 821 N. What is the acceleration

FromTheMoon [43]

Answer:

Acceleration of the elevator = 13.242 m/s2.

Explanation:

Force is defined as the push or pull on an object with mass in kg that causes a change velocity. Force as a vector means it has both magnitude and direction.

Mathematically,

F = M*a

Where

F = force in Newton.

M = mass of the object in kg.

a = acceleration due to gravity in 9.81 m/s2.

Acceleration of the elevator = 821/62

= 13.242 m/s2.

5 0

3 years ago

Read 2 more answers

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road, (b) on an uphill road?

xeze [42]

The net force on the car for either case will be zero, since the car is moving at a constant velocity.

<h3>Net force on the car</h3>

The net force on the car is the sum of all the forces acting on the car.

∑F = ma

<h3>When the car is on a level road</h3>

When the car is on a level road, the only two forces acting on the car is the applied force and frictional force of the road.

F - Ff = 0

The net force will be zero, since the car is moving at a constant velocity.

<h3>When the car is on uphill</h3>

The forces acting on the car at the uphill is the applied force, weight of the car acting downwards and the frictional force acting opposite direction.

F - Wsinθ - μFₙcosθ = 0

The net force will be zero, since the car is moving at a constant velocity.

Learn more about net force at constant velocity here: brainly.com/question/14392124

6 0

2 years ago

A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t

OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

$(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s$

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

$s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s$

Now, the distance covered by the player in this time will be:

$s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)$

<u>s₁ = 0.022 m</u>

4 0

3 years ago

A 1200-kg station wagon is moving along a straight highway at Another car, with mass 1800 kg and speed has its center of mass 40

taurus [48]

Answer:

Center of mass lies 24 m in front of center of mass of second wagon.

Explanation:

Suppose A 1200 kg station wagon is moving along a straight highway at 12.0 m/s. Another car with mass 1800 kg and speed 20.0 m/s.

Given that,

Mass of first wagon = 1200 kg

Mass of second wagon = 180 kg

Distance = 40 m

We need to calculate the position of the center of mass of the system

Using formula of center mass

$x_{cm}=\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$

$x_{cm}=\dfrac{1200\times0+1800\times40}{1200+1800}$

$x_{cm}=24\ m$

Hence, Center of mass lies 24 m in front of center of mass of second wagon.

5 0

3 years ago

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago