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2 years ago

a 50- kg block of concrete is pulled upward (not sideways) with a force of 600 n =what is the acceleration

Physics

1 answer:

Arte-miy333 [17]2 years ago

6 0

The weight of the block is (mass x gravity) = (50kg x 9.8 m/s²) = 490 N

The NET vertical force on the block is (490N down) + (600N up) = 110N up.

Acceleration = (net force) / (mass)

Acceleration = (110N up) / (50 kg)

<em>Acceleration = 2.2 m/s² up</em>

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What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

$\therefore 0=\frac{2500}{9} + a(1.1)$

$\Rightarrow a=-\frac{2500}{9(1.1)}$

$\therefore a = - 252.52 \ m/s^2$

Hence , the deceleration of the rocket is $252.52\ m/s^2$ .

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11 months ago

ou are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soo

madam [21]

Till the time car is just adjacent to the bicycle we can say

distance moved by cycle = distance moved by car

Time taken by car to accelerate from rest

$t = \frac{v_f - v_i}{a}$

$t = \frac{49 - 0}{7} = 7 s$

Time taken by cycle to accelerate

$t = \frac{23 - 0}{15} = 1.53 s$

now the distance moved by cycle in time "t"

$d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)$

distance moved by car in same time

$d = \frac{7t + 0}{2}(t)$

now make them equal

$3.5t^2 = 17.595 - 35.19 + 23t$

$3.5 t^2 - 23t + 17.595 = 0$

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