What is a Nuclei of uranium atoms split apart

A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push bo

Fynjy0 [20]

Answer:

Equal

Explanation:

The impulse theorem states that the impulse exerted on each cart is equal to the change in momentum of the cart:

$I=\Delta p=p_f -p_i$

where

I is the impulse

pf is the final momentum

pi is the initial momentum

The impulse is equal to the product between the force applied and the contact time:

$I=F\Delta t$

In this case, the force applied to the two carts (F) is the same, and the contact time ( $\Delta t$ ) is the same as well. Therefore, the impulse exerted on the two carts is the same.

Moreover, the initial momentum of the two carts is also the same (zero, because they start from rest: $p_i=0$ ). So the formula becomes

$p_f = I$

And since I is the same for the two carts, the final momentum ( $p_f$ ) will also be equal.

7 0

3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago

(I speak portuguese, but I will make a question) Se eu estou num carro, com velocidade constante de 100km/h e há uma abelha dent

denis23 [38]

Yes the bee is flying with the car at 100km/h

8 0

4 years ago

Read 2 more answers

A curved, transparent object used to refract light

zlopas [31]

It's called a concave lens.

6 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago