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3 years ago

A physics student pushed a 50 kg load across the floor, accelerating it at a rate of 1.5 m/s . How much force did she apply?

Physics

1 answer:

garri49 [273]3 years ago

4 0

Explanation:

if you're trying to find the acceleration then it's change in velocity ÷ time taken

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An hydrogen molecule consists of two hydrogen atoms whose total mass is 3.3×10−27 kg and whose moment of inertia about an axis p

dlinn [17]

Answer:

$6.9631\times 10^{-11}\ m$

Explanation:

I = Moment of inertia = $4\times 10^{-48}\ kg m^2$

m = Mass of two atoms = 2m = $3.3\times 10^{-27}\ kg$

r = distance between axis and rotation mass

Moment of inertia of the system is given by

$I=mr^2\\\Rightarrow I=2mr^2\\\Rightarrow 4\times 10^{-48}=3.3\times 10^{-27}\times r^2\\\Rightarrow r=\sqrt{\frac{4\times 10^{-48}}{3.3\times 10^{-27}}}\\\Rightarrow r=3.48155\times 10^{-11}\ m$

The distance between the atoms will be two times the distance between axis and rotation mass.

$d=2r\\\Rightarrow d=2\times 3.48155\times 10^{-11}\\\Rightarrow d=6.9631\times 10^{-11}\ m$

Therefore, the distance between the two atoms is $6.9631\times 10^{-11}\ m$

3 0

4 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic

valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring with respect to its equilibrium position

For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(84.08)}{342.25}}=0.701 m$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0

3 years ago

Landon's new toy arrived in the mail covered in foam packing peanuts. The foam packing peanuts have a mass of 30 g and a volume

gogolik [260]

The formula for calculating density is P=M/V where P is the density, M is the mass, and V is the volume.

The problem gives you the mass, 30g, and the volume, 60cm^3;you can plug those into the equation, which should give you P=30/60.

Your answer should end up being P=0.5 g/cm^3.

WORK:

P=M/V

P=30g/60cm^3

P=0.5g/cm^3

4 0

3 years ago

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

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

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

8 0

3 years ago