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3 years ago

An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

Physics

1 answer:

zmey [24]3 years ago

6 0

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

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3 years ago

Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg

jolli1 [7]

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

Kinetic energy of the bowling ball,

$E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J$

The kinetic energy of the bocce ball,

$E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J$

So, the kinetic energy of bocce ball is more than that of bowling ball.

5 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3

ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0

3 years ago