Ask question

Ask question

All categories

3 years ago

15

A box that is sliding across the floor experiences a net force of 10.0 N. If the box has a mass of 1.50 kg, what is the resultin

g acceleration of the box g

1 answer:

Maksim231197 [3]3 years ago

7 0

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

You might be interested in

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

DerKrebs [107]

The ball's horizontal and vertical velocities at time $t$ are

$v_x=v_{xi}$

$v_y=v_{yi}-gt$

but the ball is thrown horizontally, so $v_{yi}=0$ . Its horizontal and vertical positions at time $t$ are

$x=v_{xi}t$

$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

3 years ago

Car a is heading north with a velocity of 14 m/s. car b is traveling north also, with a velocity of 19 m/s. what is the relative

kicyunya [14]

a.) Relative velocity = velocity of car B - velocity of car A
= 19 - 14
= 5 m/s
Hope this cleared your doubt, feel free to ask any question regarding this and have a nice day ahead! :)

6 0

4 years ago

Read 2 more answers

In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?

yawa3891 [41]

Answer:

actually I was just wondering what you are thinking

6 0

3 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?

AleksandrR [38]

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

WD= 45*6

WD=270Joues

7 0

3 years ago