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3 years ago

Helppp pliss.............

Physics

1 answer:

Serggg [28]3 years ago

8 0

i don't know sorry

free to comment if you know the answer

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Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h

vladimir1956 [14]

Answer:

Explanation:

Given

Mass of block A $(m_a)$ =2.6 kg

Mass of block B $(m_b)$ =0.81 kg

Force of 6.5 N is applied on Block A

Force on Block A is F, Tension

thus

$F-T=m_a\times a$ -----1

where a=acceleration of the system

For Block B

$T=m_b\times a$ -----2

$F=m_b\times a+m_a\times a$

$F=\left ( m_a+m_b\right )a$

$F=\left ( 2.6+0.81\right )a$

$a=\frac{6.5}{3.41}$

$a=1.96 m/s^2$

$T=0.81\times 1.96=1.54 N$

5 0

3 years ago

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

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

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

A wave of frequency 500hz and wave length 50m is traveling into a medium, dertimine the speed of velocity of the wave. formula V

ololo11 [35]

Answer:

25000m/s

Explanation:

V=fd

V=(500)(50)

V=25000m/s

6 0

3 years ago

Read 2 more answers

Two objects of different masses have momentum of equal, non-zero magnitude. Which object has more kinetic energy?.

Semmy [17]

If they have the same momentum, then

m₁ v₁ = m₂ v₂

If m₁ > m₂, then we must have v₁ < v₂ to preserve the equality.

The object with the larger speed - the second object - thus has more kinetic energy.

8 0

2 years ago

A vehicle is moving at a speed of 20m/s suddenly the driver sees a fallen tree on the road, he instantly applies the brakes that

NeTakaya

Answer:

b) Distance covered before stopping.

Explanation:

If you want to find the distance d required to come to a stop starting at some initial speed v, with braking acceleration a, use the kinematic relation

vf2 - vi2 = 2ad

vi = initial speed at the moment braking begins = v

vf = final speed = 0 (comes to a full stop)

-v2 = 2ad

d = -v2/(2a)

6 0

3 years ago