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3 years ago

According to Newton's first law, a moving odject acted on by a net force of zero-

Physics

1 answer:

avanturin [10]3 years ago

3 0

C, moves at a constant speed. since no force is acted on it, the energy it uses to go forward does not change either

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A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th

NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

$v=r\omega$

$v = 4.50\times110\times\dfrac{2\pi}{60}$

$v=51.83\ m/s$

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0

3 years ago

Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e

anyanavicka [17]

Answer:

Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

F + F = m a

k₃ x + k₃ x = m a

a = 2k₃ x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

Keq = 2k₃

6 0

3 years ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

A 60 Watt light bulb runs for 120 seconds. How much energy does it use?

sattari [20]

ENERGY = POWER X TIME
=60 X 120=7200KWh

6 0

3 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago