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3 years ago

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A truck with a mass of 5000 kg and transfers 800,000 newtons(N) of force when it hits a wall. What was the acceleration of the t

ruck?
a. 160 m/s
b. 160 km/hr
c. 160 m/s2
d. 160 miles/hr

1 answer:

Troyanec [42]3 years ago

4 0

Answer is C. 160 m/s2

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Please help with 3 questions about acceleration

Goryan [66]

A great, helpful, useful definition of acceleration is

<em>A = (change in speed) / (time for the change)</em> . <== you should memorize this

This simple tool will directly solve all 3 problems.

The REASON for assigning these problems for homework is NOT to find the answers. It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do. Noticed that if you get the answers from somebody else, you lose all of these benefits.

The only wrinkle anywhere here is in #3, because when you use this definition, the unit of time has to be the same in both the numerator and the denominator.

So for #3, you have to EITHER change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.

5 0

4 years ago

A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them

Gwar [14]

Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

5 0

3 years ago

Read 2 more answers

On a cross-country trip, a couple drives 500 mi in 10 h on the first day, 380 mi in 8.0 h on the second day, and 600 mi in 15 h

Zepler [3.9K]

We know that the average speed is simply the ratio of the total distance travelled over the total duration of the trip.

total distance = 500 mi + 380 mi + 600 mi

total distance = 1,480 mi

total time = 10 h + 8 h + 15 h

total time = 33 h

So the average speed is therefore:

average speed = 1,480 mi / 33 h

<span>average speed = 44.85 mi / h</span>

8 0

3 years ago

What is the definition of Rock Strata and Law of Original Horizontality?

timurjin [86]

It's called the <span>Principle of Original Horizontality
</span><span> it just </span>means<span> what it sounds like: that all </span>rock layers <span>were originally horizontal.
</span>Of course, it only applies to sedimentary rocks<span>.
</span>Recall that sedimentary rock is composed of <span> sediments, which are deposited and compacted in one place over time.</span>

3 0

3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago