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2 years ago

6

a student stands on a metric scale, his weight is 490 N. then another student hands him a 5.00-kg mass to hold. the force shown

on the scale is now

1 answer:

Gwar [14]2 years ago

4 0

Answer:

539.04 N

Explanation:

5kg is approximately 49.04 newton's so add 49.04 to 490 and you get 539.04 N

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The horizontal surface on which the objects slide is frictionless. If M = 1.0 kg and the magnitude of the force of the small blo

sweet-ann [11.9K]

Answer:

The force is 7.8 N.

Explanation:

Given that,

Mass of small object = 2M

Mass of large object = 4M

Here, M = 1.0 kg

Force of the small block = 5.2 N

We need to calculate the acceleration of 4 kg block

Using formula of force

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{5.2}{4}$

$a=1.3\ m/s^2$

The 2 kg block is also accelerating at 1.3 m/s², making a total of 6 kg.

We need to calculate the force

Using formula of force

$F=ma$

Put the value into the formula

$F=6\times1.3$

$F=7.8\ N$

Hence, The force is 7.8 N.

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3 years ago

A golf ball is dropped from rest from a height of 9.50m. It hits the pavement, then bounces back up, rising just 5.70m before fa

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Answer:

Explanation:

Given

height from which ball is dropped $h_1=9.5\ m$

and it rises to a height of $h_2=5.7\ m$

A person caught the ball in mid-way i.e. at a height of $h_0=1.2\ m$ from pavement

time taken to reach the bottom is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

here initial velocity is zero i.e.

$h_1=0+\frac{1}{2}gt_1^2$

$t_1=\sqrt{\frac{2h_1}{g}}$

Similarly to reach of height $h_2$ time taken is $t_2$

$t_2=\sqrt{\frac{2h_2}{g}}$

The ball is caught in mid-way i.e. ball travel a distance of $h_3=5.7-1.2=4.5\ m$

time taken is $t_3=\sqrt{\frac{2h_3}{g}}$

total time taken $t=t_1+t_2+t_3$

$t=\sqrt{\frac{2h_1}{g}}+\sqrt{\frac{2h_2}{g}}+\sqrt{\frac{2h_3}{g}}$

$t=\sqrt{\frac{2\cdot 9.5}{9.8}}+\sqrt{\frac{2\cdot 5.7}{9.8}}+\sqrt{\frac{2\cdot 4.5}{9.8}}$

$t=3.429\ s$

total time =3.429 s

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3 years ago