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evablogger [386]

3 years ago

5

An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (cons

tant) acceleration of the vehicle during this time? Group of answer choices

1 answer:

Nata [24]3 years ago

8 0

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

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he electric motor of a model train accelerates the train from rest to 0.720 m/s in 29.0 ms. The total mass of the train is 840 g

kkurt [141]

Answer:

7.481W

Explanation:

power= work done÷time

power= force × uniform velocity

a= ∆v/t

29.0ms=

$29 \times {10}^{ - 3 } s$

=(0.72-0)÷29×10^-3

a=24.83m/s^2

Newton's second law

f=ma

=(0.84×24.83)

=20.86N

w=FD

v^2=u^2+2as

0.5184=0+49.66s

s=0.0104m

20.86×0.0104

=0.216944J

0.216944÷t

P=7.481W

4 0

2 years ago

A crate is given a push across a horizontal surface. The crate has a mass m, the push gives it an initial speed of 1.90 m/s, and

Roman55 [17]

Answer:

a) $s = 1.534\,m$ , b) $s = 6.135\,m$

Explanation:

a) The energy equation for the crate is modelled after the Principle of Energy Conservation and Work-Energy Theorem. Changes in gravitational potential energy can be neglected due to the information of a horizontal surface:

$K_{A} = W_{loss}$

$\frac{1}{2}\cdot m \cdot v^{2} = \mu_{k} \cdot m \cdot g\cdot s$

The distance that crate needs to cover before stopping is:

$s = \frac{v^{2}}{2\cdot \mu_{k}\cdot g}$

$s = \frac{(1.90\,\frac{m}{s} )^{2}}{2\cdot (0.120)\cdot (9.807\,\frac{m}{s^{2}} )}$

$s = 1.534\,m$

b) The stopping distance is:

$s = \frac{(3.80\,\frac{m}{s} )^{2}}{2\cdot (0.120)\cdot (9.807\,\frac{m}{s^{2}} )}$

$s = 6.135\,m$

3 0

3 years ago

Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef

Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μ $F_n$ where μ is the coefficient of friction, and $F_n$ is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

$F_n=mg$ so

$F_n=(105)(9.8)$ so the weight of the sled is

$F_n=$ 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0

3 years ago

A 5 Kg bowling ball is thrown at a stationary 1.6 Kg bowling pin at 5 m/s. If the final velocity of the ball is 2.5 m/s. The fin

Sergio039 [100]

Answer: hope this helps

Explanation:

7 0

3 years ago

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

3 years ago