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STALIN [3.7K]
1 year ago
10

The standard acceleration (at sea level and 45◦ latitude) due to gravity is 9.806 65 m/s2. What is the force needed to hold a ma

ss of 2 kg at rest in this gravitational field? How much mass can a force of 1 N support?
Physics
1 answer:
erma4kov [3.2K]1 year ago
5 0

a) Remember Newton's second Law:

F = m*a

This means that the forces that is being exerted over an object is equal to the mass of the object times the acceleration that it has.

In this case, in order to hold the mass you need a force with the same magnitude but opposite direction to the gravitational force. The magnitude of this force would be the mass of the object times the gravitactional  acceleration:

F = 2kg*9.80665m/s² = 19.6133 N

b) Using again Newton's second Law, we can issolate mass from that equation:

m = F/a

Then, the mass that a force of 1N can support is equal to:

m = 1N/9.80665m/s² = 0.102 kg

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1 year ago
At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m
Margaret [11]

Answer:

Part a)

t = \sqrt{\frac{2h}{g}}

Part b)

t = 1.06 s

Part c)

L  = 4.86 m

Explanation:

Part a)

The height of the diving board is given as

h = 5.5 m

now the speed of the diver is given as

v_0 = 2.7 m/s

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

y = v_y t + \frac{1}{2}at^2

h = 0 + \frac{1}{2}gt^2

t = \sqrt{\frac{2h}{g}}

Part b)

t = \sqrt{\frac{2h}{g}}

plug in the values in the above equation

t = \sqrt{\frac{2(5.5 m)}{9.81}

t = 1.06 s

Part c)

Horizontal distance moved by the diver is given as

d = v_0 t

d = 2.7 \times 1.06

d = 2.86 m

so the distance from the edge of the pool is given as

L = 2.86 + 2

L  = 4.86 m

4 0
1 year ago
A Styrofoam slab has thickness h and density ρs. When a swimmer of mass m is resting on it, the slab floats in fresh water with
sattari [20]
 <span>A = area of styrofoam 
M = mass of stryofoam = A*h*rho_s 
m = mass of swimmer 

Total mass = m + M = m + A*h*rho_s 
Downward force = g*(total mass) = g*[m + A*h*rho_s] 

The slab is completely submerged. 
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w] 

Equate these 
g*[m + A*h*rho_s] = g*[A*h*rho_w] 
m + A*h*rho_s = A*h*rho_w 
A*h*[rho_w - rho_s] = m 
A = m/[h*(rho_w - rho_s)]</span>
8 0
1 year ago
Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
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4 0
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