All categories

2 years ago

A wooden block slides down a ramp that is 12 meters long in 4 seconds. What was the average speed of the wooden block?\

Physics

2 answers:

Sholpan [36]2 years ago

4 0

speed=distance/ timetaken

ans:

12m/4s

=3m/s

ipn [44]2 years ago

3 0

The answer is B) 3 m/s

Mark as brainliest ^-^

You might be interested in

The mass of a coin is measured to be 12.5±0.1 g. The diameter is 2.8±0.1 cm and the thickness 2.1 ±0.1 mm. Calculate the average

Evgesh-ka [11]

The average density of the material from which the coin is made is 9.67 g/cm³.

<h3>Volume of the coin</h3>

The volume of the coin at the given diameter is calculated as follows;

V = Ah

where;

A is area of the coin
h is the thickness of the coin

V = πd²/4 x h

V = π(2.8)²/4 x (0.21 cm)

V = 1.293 cm³

<h3>average density of the coin</h3>

The average density of the material from which the coin is made is calculated as follows;

density = mass/volume

density = 12.5 g / (1.293 cm³)

density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

Learn more about average density here: brainly.com/question/1354972

#SPJ1

7 0

1 year ago

WILL GIVE BRAINLIEST

NemiM [27]

Yes! you are :) bc you are FORCING the page to turn, and the broom ti sweep

6 0

3 years ago

Read 2 more answers

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

$E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta$

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

$\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0$

4 0

3 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

3 years ago

A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature $\left ( T_H\right )=240^{\circ}\approx 513$

Lower Temprature $\left ( T_L\right )=20^{\circ}\approx 293$

Engine power ouput $\left ( W\right )=910 W$

Efficiency of carnot cycle is given by

$\eta =1-\frac{T_L}{T_H}$

$\eta =\frac{W_s}{Q_s}$

$1-\frac{293}{513}=\frac{910}{Q_s}$

$Q_s=2121.954 W$

$Q_r=1211.954 W$

rounding off to two significant figures

$Q_r=1200 W$

5 0

3 years ago