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3 years ago

A car moving on a straight level road changes its speed from 3.5m/s to 4.5m/s in 5.00 seconds. Find it’s acceleration

Physics

1 answer:

mixer [17]3 years ago

8 0

Answer:

0.2m/s2

Explanation:

acceleration= change in velocity ( in this case speed)/ change in time.

4.5-3.5=1m/s

change in time=5-0 =5

1m/s divided by 5 seconds acceleration as the result

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You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x di

Volgvan

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

$f = \mu R = \mu W$

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\$

8 0

3 years ago

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$ . So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$ , s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$ . So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

5 0

3 years ago

A cannonball is launched from the ground with 48279 Joules of kinetic energy. How much gravitational potential energy will the c

Mamont248 [21]

When the cannonball is shot upwards, kinetic energy converts to gravitational potential energy (GPE).

As the ball rises, speed decreases and height increases.

So when the cannonball has reached its maximum height, all of the Kinetic energy has transferred into gravitational potential energy.

(Because at the max height, the cannonball for a brief moment has no velocity, and thus no kinetic energy)

So the GPE is 48279 Joules at it's maximum height.

6 0

4 years ago

Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido

almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

$v_{P/B} = v_{P} - v_{B}$ (1)

Donde:

$v_{P/B}$ - Velocidad del perro relativo al barco, en kilómetros por hora.

$v_{P}$ - Velocidad del perro con respecto al observador, en kilómetros por hora.

$v_{B}$ - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que $v_{B} = 30\,\frac{km}{h}$ y $v_{P/B} = 10\,\frac{km}{h}$ , entonces la velocidad del perro con respecto al observador es:

$v_{P} = v_{B} + v_{P/B}$

$v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}$

$v_{P} = 40\,\frac{km}{h}$

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0

3 years ago

The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl

aleksandr82 [10.1K]

$acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}$

6 0

4 years ago