All categories

3 years ago

What force is needed to accelerate a 0.5kg football at a rate of 40m/s

Physics

1 answer:

GREYUIT [131]3 years ago

6 0

Force=mass x acceleration
f= 0.5 x40
f=20N

You might be interested in

A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance

ArbitrLikvidat [17]

Answer:

The value is $v_y = -48.61 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $u_x = 252 \ m/s$

The horizontal distance is $d = 1250 \ m$

Generally the time taken by the hot magma in air before landing is mathematically represented as

$t = \frac{d}{u_x}$

=> $t = \frac{ 1250 }{252}$

=> $t = 4.96 \ s$

Generally the initial vertical velocity of the magma when it was lunched is

$u_y = 0 \ m/ s$

Then the final velocity of the magma when it hits the ground is mathematically represented s

$- v_y = u_y + gt$

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

$- v_y = 48.61 \ m/s$

=> $v_y = -48.61 \ m/s$

7 0

2 years ago

CAN SOMEONE PLEASE HELP ME ASAP?

ICE Princess25 [194]

The first one is dependent variable
<span />

7 0

3 years ago

Read 2 more answers

Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po

Brrunno [24]

Answer:

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across $R_1=\Delta V_1$

Potential difference across $R_2=\Delta V_2$

Potential difference across $R_3=\Delta V_3$

We know that in series combination

Potential difference , $V=V_1+V_2+V_3$

Using the formula

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Hence, this is required expression for potential difference.

3 0

3 years ago

What is a transfer a wave

makkiz [27]

Energy is transferred in a wave

Energy is transferred, but mass is not.

4 0

3 years ago

Read 2 more answers

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers