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1 year ago

Compare the density, weight, mass, and volume of a pound of gold to a pound of iron on the surface of Earth.

Physics

1 answer:

Nitella [24]1 year ago

8 0

Answer:

Mass of gold = mass of iron

weight of gold = weight of iron

volume of gold is less than the volume of iron

Explanation:

mass of gold = 1 pound

mass of iron = 1 pound

Mass of both the metals is same.

Weight is equal to the product of mass and the acceleration due to gravity. As acceleration due to gravity is same for both the metals so the weight of both the metals is same.

As the density of iron is less than the density of gold and we know that the volume is defined as the ratio of mass to the density.

mass is same for both the metals, so the volume of iron is more than the density of gold.

Mass of gold = mass of iron

weight of gold = weight of iron

volume of gold is less than the volume of iron

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When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup

Wittaler [7]

Answer:

$W_f = 148.17J$

Explanation:

During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.

Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

$N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N$

The work done is given by the friction force and the distance traveled,

$W_f = fd = \mu_k Nd$

Where $\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)$

The thermal energy released through the work done is,

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2 years ago

Question #2

Contact [7]

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

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2 years ago

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$