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Allushta [10]
3 years ago
7

A pickup truck is carrying a 10.0 kg toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts mo

ving. The coefficients of static and kinetic friction between the toolbox and the truck bed are 0.31 and 0.18, respectively. Assume that the truck bed is horizontal. What is the shortest time the truck could take to accelerate from rest to 13.2 m/s without causing the toolbox to slide

Physics
1 answer:
scoray [572]3 years ago
5 0

Answer: The shortest time the truck could take is 4.34 secs

Explanation: Please see the attachments below

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A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

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

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67
aliya0001 [1]
Acceleration =velocity /time
=42/7
=6
5 0
3 years ago
Read 2 more answers
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
3 years ago
Read 2 more answers
What happens when a force is applied to an object in stable equilibrium
Katarina [22]

Answer:

the object is no longer in equilibrium .

7 0
3 years ago
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in
dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860  m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}

Put all the values,

t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s

The velocity of the ball is :

v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s

The kinetic energy of the ball is :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J

So, the required kinetic energy is 0.076 J.

6 0
3 years ago
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