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3 years ago

What is the average velocity of the object between 14 and 22 seconds

Physics

2 answers:

anastassius [24]3 years ago

4 0

Use the slope equation. It’s 1-8 Meters which is 7. And it’s 22-14 seconds. So the answer is d

Natalka [10]3 years ago

4 0

Answer:

v = - 0.88 m/s

Explanation:

Average velocity is defined as the ratio of displacement and time

so here as we can see that the displacement is the change in the position of the object

initially object is placed at x = 8 m

final position of the object is x = 1

so the displacement is given as

$d = x_f - x_i$

$d = 1 - 8 = -7 m$

now the time taken by the object

$\Delta t = 22 - 14 = 8 s$

now by the formula of average velocity we know

$v = \frac{displacement}{time}$

$v = \frac{-7}{8} = - 0.88 m/s$

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1 year ago

A car travels 460km in 7 hours. How would you calculate the speed?

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2 years ago

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

2 years ago

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

We know that

F = m a = E q ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0

3 years ago

Read 2 more answers

PLZ HELP

lianna [129]

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2 years ago