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skelet666 [1.2K]
3 years ago
7

A 3.50 kg block is pushed along a horizontal floor by a force of magnitude 15.0 N at an angle θ = 30.0° with the horizontal. The

coefficient of kinetic friction between the block and the floor is 0.250.
Calculate the magnitudes of (a) the frictional force on the block from the floor and
(b) the block's acceleration.
Physics
1 answer:
Crank3 years ago
8 0

Answer :

The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

Explanation :

Given that,

Mass of block = 3.50

Angle = 30°

Force = 15.0 N

Coefficient of kinetic friction = 0.250

We need to calculate the frictional force

Using formula of frictional force

F_{k}=\mu N

F_{k}=\mu (F\sin\theta+mg)

F_{k}=0.250\times(15\times\sin30^{\circ}+3.50\times9.8)

F_{k}=0.250\times41.8

F_{k}=10.45\ N

(II). We need to calculate the block's acceleration

Using newton's second law of motion

F=ma

a=\dfrac{F}{m}

a=\dfrac{F\cos\theta-F_{k}}{m}

a=\dfrac{15.0\cos30^{\circ}-10.45}{3.50}

a=0.73\ m/s^2

Hence, The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

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Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = -2 \times 10^2 C

q_2 = -4 \times 10^{-8} C

r = 3.0 m

now we have

F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}

F = 8000 N

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

3 0
3 years ago
A ball is let go and falls to the ground. It bounces a few times before
Ksenya-84 [330]

Answer:

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4 0
3 years ago
A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
Inessa05 [86]

Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of

<em>F</em> = <em>m a</em>   →   <em>a</em> = <em>F </em>/ <em>m</em>

so that the cart's final speed is

<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>

3 0
3 years ago
A ball is thrown horizontally from the top of a building 22.3 m high. The ball strikes the ground at a point 127 m from the base
DochEvi [55]

Answer:

t = 2.13 s

Explanation:

given,

height of the building = 22.3 m

horizontal distance = 127 m

acceleration due to gravity = 9.8 m/s²

time for which ball is in motion = ?

using equation of motion

s = u t + \dfrac{1}{2}gt^2

initial velocity is zero

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 22.3}{9.8}}

       t = √4.551

        t = 2.13 s

8 0
3 years ago
A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and
docker41 [41]

Answer:

T = 451.26 N

Explanation:

It is given that,

The mass of block, m = 46 kg

Mass of the chain, m' = 19 kg

Length of the chain, l = 1.9 m

Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.

T=mg

T=46\ kg\times 9.81\ m/s^2

T = 451.26 N

So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.

4 0
3 years ago
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