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Ivanshal [37]
3 years ago
5

Maisie drags a 1kg mass along a table with a Newton meter so that it accelerates at 0.25 m/s. The Newton-meter reads 0.4N, calcu

late the force of friction between the mass and the table
Physics
1 answer:
Temka [501]3 years ago
7 0
Applying one newton of force to an object of 1 kg accelerates it by 1 meter per second

maisie applied .4 newtons to the block, so it should be accelerating at .4 m/s
but it actually accelerated at .25 m/s or .25 newtons reached the block

the difference in force applied and force received is .15, which is the force of friction because that's how much force it stopped from reaching the block
force of friction is .15 newtons
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elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi
Citrus2011 [14]

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

6 0
2 years ago
1. A car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0 second time interval. What is its average acceleration?
Finger [1]

Explanation:

36-4/4= 9 m/squared. meter per squared.

acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)

7 0
2 years ago
A 73.9 kg weight-watcher wishes to climb a
Tresset [83]

The height to which the weight-watcher must climb to work off the equivalent 991 (food) Calories is 0.59 Km

<h3>How to determine the energy. </h3>

1 food calorie = 103 calories

Therefore,

991 food calories = 991 × 103

991 food calories = 102073 calories

Multiply by 4.2 to express in joule (J)

991 food calories = 102073 × 4.2

991 food calories = 428706.6 J

<h3>How to determine the height </h3>
  • Energy (E) = 428706.6 J
  • Mass (m) = 73.9 kg
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Height (h) =?

E = mgh

Divide both side by mg

h = E / mg

h = 428706.6 / (73.9 × 9.8)

h = 591.95 m

Divide by 1000 to express in km

h = 591.95 / 1000

h = 0.59 Km

Learn more about energy:

brainly.com/question/10703928

7 0
2 years ago
Final velocity will be greater than initial velocity of an object is
Natali [406]

Answer:

accelerating

Explanation:

If we consider(v > u) Acceleration:

final velocity(v)= 14m/s

initial velocity(u)=10m/s

time taken(t)= 2 seconds

a= \frac{(v-u)}{t} =\frac{(14-10)}{2}=2m/s²

If we consider (v<u) Deceleration:

final velocity(v)= 3m/s

initial velocity(u)=9m/s

time taken(t)=2 seconds

a= \frac{(v-u)}{t}=\frac{(3-9)}{2}= -3m/s²

4 0
2 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
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