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3 years ago

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Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with

a length of 9.3m

1 answer:

NARA [144]3 years ago

7 0

1.5 m/s is the velocity. 9.3 m is the length of aisle, over which Distance will be covered. Time is demanded in which the child will move the cart over the aisle with 1.5 m/s. v=S/t and, t=S/v Put values, t=9.3/1.5=6.2 s

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

2 years ago

A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (b) What mass of lead should be p

schepotkina [342]

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

<h3>What mass of lead should be placed on the cube?</h3>

Given: Side of the cube (a) = 20cm

The density of the cube (ρc) = $$$650 kg/m^3$

a) Applying the force balance, the buoyant force must be equal to the weight of the cube

ρcgV = ρg × (Ax)

Substituting the values in the above equation, we get

$(650*(0.2 m)^3)=1000*(0.2 m)^2*x$

x = 0.13

where x is the height of the cube in the water

$$$A = a^2$ is the area of the cross-section

ρ is the density of the water

V is the volume of the cube

Now, the height above the surface of the water would be

h = a − x

Substituting the values, then we get

h = 0.2 − 0.13

h = 0.07 m

b) The mass added is "m" so the complete cube is submerged in the water, therefore

ρcgV + mg = ρg × (V)

$$(650*(0.2 m)^3)+m=1000*(0.2 m)^3$

m = 2.8 kg

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

To learn more about buoyant force refer to:

brainly.com/question/11884584

#SPJ4

7 0

1 year ago

Umnica [9.8K]

Since the baseball is half the weight of the softball, the baseball has to go twice as fast as the softball.

So the best answer would be D

5 0

3 years ago

Read 2 more answers

___________ is the process by which wind removes surface materials.

Travka [436]

The answer is deflation...have a good day

4 0

3 years ago

Read 2 more answers

We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap

marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B= $3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm= $0.5\times 10^{-3} m$

Radius of wire,r= $\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

$1 cm=10^{-2} m$

Resistivity of copper, $\rho=1.68\times 10^{-8}\Omega m$

We know that

$R=\frac{\rho l}{A}$

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm

8 0

3 years ago