Whenever an object is standing still , which value is always zero?

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

Zinaida [17]

Answer:

4.17 m/s²

Explanation:

We are told the reaction time is 0.2 s. Now, during this reaction time the car is going to travel an additional distance of : x = u × t = 40 × 0.2 = 8 m

where u is the initial velocity of the car which is 40.0 m/s.

We are told that he had 200 m to stop before applying brakes. Thus, after applying brakes, he now has a distance to cover of; s = 200 - 8 = 192 m

Since vehicle is coming to rest acceleration would be negative, thus using Newton's equation of motion, we have;

v ² = u² - 2as

v = 0 m/s since it's coming to rest

u = 40 m/s

s = 192 m

Thus;

0² = 40² - 2(a)(192)

0² = 1600 - 384a

a = 1600/384

a = 4.17 m/s²

6 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

A water pump is a positive displacement-type pump true or false

Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0

4 years ago

A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

vazorg [7]

Answer:

8z

Explanation:

It is 8z

3 0

4 years ago

What are all the moon phases

kicyunya [14]

New Moon
Waxing Crescent
First Quarter
<span>Waxing Gibbous
</span>Full Moon
<span>Waning Gibbous
</span>Last Quarter
Waning Crescent

;)

8 0

4 years ago

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