The engine of a jet airplane pushes exhaust gases from burning fuel backward

A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets th

SashulF [63]

Answer:

Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J

Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s = ¹/₂×at²

$t = \sqrt{\frac{2s}{a} }$

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

$t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s$

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

= ¹/₂ × 2.5× (4.531)²

= 25.663 J

6 0

4 years ago

Poate cineva sa imi rezolve fisa Asta pana Maine?

Firlakuza [10]

Answer:

httfytstuytshysts

Explanation:htrhstrhhstrsstrthrssr

5 0

3 years ago

How long must a simple pendulum be if it is to make exactly one swing per second? (That is, one complete vibration takes exactly

umka2103 [35]

Oscillation is a type of periodic motion which repeats itself to and from about a point which is called mean position. The period of a Pendulum can be described as a ratio between the length and gravity as,

$T = 2\pi \sqrt{\frac{L}{g}}$

Here,

L = length

g = Gravity

If we rearrange to find the length we have that,

$L = \frac{T^2g}{4\pi^2}$

Our period is 2s and the gravity is 9.8m/s^2, then,

$L = \frac{(2)^2(9.8)}{4\pi^2}$

$L = 0.9929m$

The simple required length of the pendulum must be 0.9929m

8 0

4 years ago

A block of mass 1.3 kg is resting at the base of a frictionless ramp. A bullet of mass 50 g is traveling parallel to the ramp su

aleksandrvk [35]

Answer:

s = 2.65 m

Explanation:

given,

mass of block,M = 1.3 Kg

mass of bullet,m = 50 g = 0.05 Kg

speed of bullet,u = 250 m/s

speed of bullet after collision,v = 100 m/s

distance traveled by the block = ?

Assuming the angle of inclination of ramp equal to 40°

calculating the speed of the block

using conservation of momentum

M u' + m u = m v + M v'

initial speed of the block is equal to zero

0 + 0.05 Kg x 250 = 0.05 x 100 + 1.3 x v'

1.3 v' = 7.5

v' = 5.77 m/s

now, calculation of acceleration

equation the horizontal component

-mg sin θ = ma

a = - g sin θ

a = - 9.8 x sin 40°

a = -6.29 m/s²

using equation of motion for the calculation of distance moved by the block

v² = u² + 2 a s

0² = 5.77² + 2 x (-6.29) x s

12.58 s = 33.29

s = 2.65 m

hence, the distance moved by the block is equal to 2.65 m

3 0

4 years ago

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

Umnica [9.8K]

Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

7 0

3 years ago