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postnew [5]
3 years ago
14

A car accelerates from rest at 2 m/s. what is the speed after 8 sec?

Physics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

16m/s

Explanation:

v_{f}=v_{i}+at

v_{f}=0+2\cdot8

v_{f}=16\ \frac{m}{s}

Therefore,  the speed after 8 seconds is 16m/s

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A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of
DENIUS [597]

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × 10^{-3}  kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 ×10^{-2} m

to find out

the speed of the mass

solution

we will apply here conservation energy  that is

K.E + P.E = Total energy  ..................1

so that Total energy = K.E max = P.E max

we know  amplitude so we find out first P.E max that is  

PE max = K.E + P.E  

(1/2)kA² = (1/2)mv² + (1/2)kx²  

kA^² =  mv²+ kx²

so here v²  will be

v²  = k(A² - x²) / m  

v = √[(k/m)×(A² - x²)]  ............2

here x = (1/2)A   so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × 10^{-3} )×(3/4×(38.9 ×10^{-2})²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s

6 0
3 years ago
5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

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3 years ago
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What is the wavelength of a bullet that is 0.450 g traveling at 2000. m/s?
ziro4ka [17]
Λ= 7.36E - 34 m

Use the Broglie wavelength equation. 

λ= { h/ mv} * SQRT {1- (v^2/ c^2)}
7 0
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