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3 years ago

Help please, It's for science :>

Physics

2 answers:

tekilochka [14]3 years ago

8 0

Help on what I don’t so anything?

xenn [34]3 years ago

3 0

Answer:

he tail of the arrow moves a distance of 0.5 m as the arrow is shot. yare yare daze

Explanation:

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8 0

3 years ago

Read 2 more answers

Fredrick did an investigation on mass and acceleration. He applied a force to a toy car and measured the distance it moved. He t

garik1379 [7]

Explanation:

B. More mass results in less acceleration.

6 0

3 years ago

(a) Find the acceleration of B.<br>(b) Find the tensions, T1 and T2, in the strings.

Ann [662]

i believ that the answer would be

the acceleration of B is 0.2

6 0

3 years ago

A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to

Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

From newton's second law:

Force = mass x acceleration

We know that acceleration is the change in velocity with time;

acceleration = $\frac{change in velocity }{time }$

Force = mass x $\frac{change in velocity }{time }$

Force = $\frac{2.964 x 7}{1.9}$ = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

3 years ago