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babymother [125]

3 years ago

9

block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal fo

rce F on the block due to the ramp

1 answer:

oee [108]3 years ago

6 0

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force

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In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.

nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

$p=mv$

$p=25kg*3m/s= 75kg*m/s$

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

$75kg*m/s = 5.0kg*v$

$v=\dfrac{75}{5}$

$\boxed{v=15m/s}$

8 0

3 years ago

Was the color orange named after the fruit or was the fruit named after the color

Marizza181 [45]

Answer:

The color orange is named after the fruit

8 0

3 years ago

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

6 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

a

$F = -1.07 *10^{-8} \ N$

b

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

So

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

3 years ago

9. a. Determine the mass of a football which has a weight of 0.80 N on a planet where the gravitational field strength is 2 N/kg

krek1111 [17]

Answer:

m = 0.4 [kg]

Explanation:

Weight is considered as a force and this is equal to the product of mass by gravitational acceleration.

$W=m*g\\$

where:

W = weight = 0.8 [N]

m = mass [kg]

g = gravity acceleration 2[N/kg]

Therefore:

$m=W/g\\m = .8/2\\m = 0.4 [kg]$

5 0

3 years ago