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3 years ago

9

The equation for distance is d = st. If a car has a speed of 35 m/s, how long will it take to go 100 m? A.2.9 s B.3500 s C.65 s

D.5.9 s

2 answers:

soldi70 [24.7K]3 years ago

6 0

I thinks that the answer is B.3500 s

aleksandrvk [35]3 years ago

3 0

D = 100
s = 35

d = st ⇒ t = d/s

Solve for t.

The answer is not B!

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A sheet of aluminum with a rectangular shape is attached to a vertical support by a set of hinges. Assume the sheet is uniform a

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3 years ago

What is the instantaneous speed at 5 seconds? (remember s=d/t)

TEA [102]

5 seconds is a poor time to ask about, because the speed abruptly changes at exactly 5 seconds.

Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.
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21. A toy car starts from rest and begins to

Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

8 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1) . After sitting a

Alekssandra [29.7K]

Answer:

0.0443 m/s

Explanation:

$m_1$ = Mass of honeybee = 0.15 g

$m_2$ = Mass of popsicle stick = 4.75 g

$v_1$ = Velocity of honeybee

$v_2$ = Velocity of stick = 0.14 cm/s

In this system the linear momentum is conserved

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The velocity of the bee is 4.43 cm/s or 0.0443 m/s

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