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sleet_krkn [62]

1 year ago

13

The weight of an object on the surface of mass is 1850 Newton and its mass is 500kg.Find the acceleration due to gravity on mars

1 answer:

Fudgin [204]1 year ago

3 0

F = m.a

a = F/m

a = 1850/500

a = 3.7 m/s²

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Question 2 (10 points)

kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

$a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2$

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

$F=ma$

where

m = 1390 kg is the mass of the truck

$a=-3.6 m/s^2$ is the acceleration

And substituting,

$F=(1390)(-3.6)=-5004 N$

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

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#LearnwithBrainly

8 0

2 years ago

The amount of energy carried by a wave and the waves maximum displacement from the rest position describe what?

Mademuasel [1]

What you are describing is Amplitude.

6 0

2 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the

saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

$R=\alpha \frac{l}{s}$

Were:

R is resistance

$\alpha$ is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area ( $s=\pi r^{2}$ ).

So if the lenght and radius are doubled, the equation goes as follows:

$R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }$

So finally because the circular area is a square function, the resulting equation is half of the one before.

7 0

3 years ago

A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

a)

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

b)

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

5 0

2 years ago

Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level $\beta _1=124 dB$

distance $r_1=5.01 m$

From sound Intensity

$\beta =10dB\log (\frac{I_1}{I_0})$

$124=10dB\log (\frac{I_1}{I_0})$

$12.4=\log (\frac{I_1}{I_0})$

$I_1=(1\times 10^{-12})\times 10^{12.4}$

$I_1=2.51 W/m^2$

we know Intensity $I\propto ^\frac{1}{r^2}$

$I_1r_1^2=I_2r_2^2$

$I_2=I_1(\frac{r_1}{r_2})^2$

$I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2$

$I_2=1.24\times 10^{-5} W/m^2$

Sound level corresponding to $I_2$

$\beta _2=10\log (\frac{I_2}{I_0})$

$\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})$

$\beta _2=70.93\approx 71 dB$

6 0

3 years ago