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3 years ago

Name the physical quantity measured by the slope of the velocity time graph for uniform motion

Physics

1 answer:

mixas84 [53]3 years ago

8 0

Acceleration. If the line is linear on a velocity time graph, the acceleration does not change, meaning its 0

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When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.

Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

$F=mg=(4.32)(9.8) = 42.3 N$

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

$x=2.92 cm = 0.0292 m$

We can find the spring constant by using Hooke's law:

$F=kx$

where k is the spring constant. Solving for k,

$k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m$

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

$F'=m'g=(1.50)(9.8)=14.7 N$

So, if we use Hooke's law again, we can find the new stretching of the spring:

$x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm$

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

$W=\frac{1}{2}kx^2$

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

$W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J$

6 0

3 years ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

Mice21 [21]

Answer:

r = 5.08 m

Explanation:

The electric force of attraction or repulsion is given by :

$F=\dfrac{kq_1q_2}{r^2}$

We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.

So, the force from the proton is balanced by the mass of the electron.

$\dfrac{kq_pq_e}{r^2}=mg$

r is distance

$r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m$

So, proton have to be at a distance of 5.08 meters above the electron.

3 0

3 years ago

If a 990 kg car is on the road and the Ff is 360 n what is the normal force

Veseljchak [2.6K]

Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.

Looking at the y-direction for the normal force:

F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons

The normal force exerted on the car by the road is 9702 newtons.

8 0

3 years ago

Read 2 more answers

The mining-resource cycle determines the use of ore materials. Which step is not part of the cycle?

ExtremeBDS [4]

Answer:

c) marketing and selling products that use ore materials

8 0

3 years ago

a car initially at rest can accelerate at 7 m/s^2 how long will it take the car to reach 60 m/s and how far will it travel durin

Alex787 [66]

1. The time taken for the car to reach a velocity of 60 m/s is 8.57 s

2. The distance travelled during the time is 257.14 m

<h3>What is acceleration? </h3>

The acceleration of an object is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

1. How to determine the time

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Time (t) =?

a = (v – u) / t

Thus,

t = (v – u) / a

t = (60 – 0) / 7

t = 8.57 s

2. How to determine the distance

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Distance (s) = ?

v² = u² + 2as

60² = 0² + (2 × 7 × s)

3600 = 0 + 14s

3600 = 14s

Divide both sides by 14

s = 3600 / 14

s = 257.14 m

Learn more about acceleration and velocity:

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7 0

1 year ago