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PSYCHO15rus [73]

3 years ago

10

Mr.smith and his wife were trying to move their new chair. Mr. Smith pulls with a force of 30N while Mrs.Smith pushes with a for

ce of 25N in the same direction. What is the net force?

1 answer:

luda_lava [24]3 years ago

7 0

Answer:

55N

Explanation:

30N + 25N = 55N

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A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra

Maslowich

Answer: $59.43\times 10^3 kg-m/s$

Explanation:

Given

mass of car $m=660 kg$

Initial velocity of car $=102 km/h\approx 28.33 m/s$ towards east

Time taken to stop $t=2.1 s$

Force exerted $F_{avg}=4.8\times 10^3 N$

change in momentum is given by impulse imparted to the car

$Impulse(J)=-F\cdot t$

$J=-4.8\times 10^3\times 2.1$

$J=-59.49\times 10^3 kg-m/s$

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum $J=59.49\times 10^3$

6 0

3 years ago

Read 2 more answers

6. A pitcher throws a ball toward home plate 18.39 meters away. If the ball is traveling at a constant 40.0 m/s, how long does i

Kruka [31]

18.39 I don’t know this

3 0

3 years ago

Which of the following is the best definition of an isotope?

almond37 [142]

Option A looks like the best definition

4 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

5 0

3 years ago

g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 Ã— 1026 W and assume that th

vagabundo [1.1K]

Answer:

The value is $N = 1.107 *10^{45 } \ photons$

Explanation:

From the question we are told that

The power output from the sun is $P_o = 4 * 10^{26} \ W$

The average wavelength of each photon is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

Generally the energy of each photon emitted is mathematically represented as

$E_c = \frac{h * c }{ \lambda }$

Here h is the Plank's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }$

=> $E_c = 3.614 *10^{-19} \ J$

Generally the number of photons emitted by the Sun in a second is mathematically represented as

$N = \frac{P }{E_c}$

=> $N = \frac{4 * 10^{26} }{3.614 *10^{-19}}$

=> $N = 1.107 *10^{45 } \ photons$

5 0

3 years ago