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3 years ago

A 70 kg student now sits in the chair and the same 35 N force is applied to the chair. What is the chair's new acceleration?

Physics

1 answer:

amid [387]3 years ago

5 0

Answer:

0.5 m/s²

Explanation:

Applying,

F = ma............... Equation 1

Where F = Force applied to the chair, m = mass of the student, a = acceleration of the chair

make a the subject of the equation

a = F/m............ Equation 2

From the question,

Given: F = 35 N, m = 70 kg

Substitute these values into equation 2

a = 35/70

a = 0.5 m/s²

Hence the new acceleration of the chair is 0.5 m/s²

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Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

RideAnS [48]

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 12.6 N

d = 15.4 m

$\theta = 15.3 degree$

so we will have

$W = 12.6 \times 15.4 cos15.3$

$W = 187.2 J$

5 0

3 years ago

Please help this is physics!!!

larisa [96]

I’m pretty sure u have it right

8 0

3 years ago

: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of

nasty-shy [4]

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).

8 0

4 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago

Helppppp pleaseeeee!!!!!!!

Lorico [155]

The first one is Water
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3 0

3 years ago

Read 2 more answers