Raising the temperature of a gas will increase its pressure if the volume of the gas

You drop some ice and notice a piece appears to slide across the kitchen floor without slowing down (until it hits the wall). wh

SVETLANKA909090 [29]

D.inertia ....................

8 0

3 years ago

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In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the

Levart [38]

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

$R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\$

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

$v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\$

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

$Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\$

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

$k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\$

k = 45.95 N/m

4 0

3 years ago

If the net force acting on an object is zero, its inertia is also zero. True or false

S_A_V [24]

Answer:

its false

Explanation:

5 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

4 years ago

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago