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xeze [42]
3 years ago
10

An object starts at rest then accelerates at a rate of 5m/s^2 for 1 second and then 2m/s^2 for 2 seconds. What is the average ac

celeration over time (t€[0,3]
Physics
1 answer:
inn [45]3 years ago
7 0

Acceleration = (change in speed) / (time for the change)

-- during the first second, the object increases its speed to

(5 m/s²) · (1 s) = 5 m/s .

-- During the next 2 seconds, the object increases its speed by

(2 m/s²) · ( s) = 4 m/s

So at the end of the whole 3 seconds, its speed is (5 m/s) + (4 m/s) = 9 m/s

-- Over the whole time, its speed has changed from zero to 9 m/s.

Acceleration = (change in speed) / (time for the change)

Acceleration = (9 m/s) / (3 sec)

<em>Acceleration = 3 m/s²</em>

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3 years ago
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Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile
Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150  =  110 miles

Displacement along the 110 North freeway = 150 - 175   = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0
3 years ago
Fill in the appropriate values for each blank as it refers to ATOM 1. The number of protons present in ATOM 1 is _________.​
wlad13 [49]
3, protons are positive and there are 3 positive atoms visible
6 0
3 years ago
A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the
Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

F=\frac{2\times0.588}{\sqrt{3} }

F = 0.679 N

6 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
3 years ago
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