On the coordinate plane below, draw the vector that goes from (0,0) to (2,3)

Se coloca una piedra de 600 g en una Honda de 50 cm y se la hace girar a una velocidad de 4 m/s. Dibuja la fuerza que ejerce la

Dima020 [189]

Answer:

Fc = 19.2 N

Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

$F_c=m\frac{v^2}{r}$

m: mass of the rock = 600g = 0.6 kg

v: tangential velocity of the Honda = 4m/s

r: radius of the Honda = 50cm = 0.5m

You replace the values of m, r and v in the equation for Fc:

$F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N$

hence, the force has a magnitude of 19.2 N

If the rock would have more mass the centripetal force would be higher

4 0

3 years ago

What are some of the landforms rovers have found on Mars that show it was once an active place?

Schach [20]

Answer:

The Curiosity rover found that ancient Mars had the right chemistry to support living microbes. Curiosity found sulfur, nitrogen, oxygen, phosphorus and carbon-- key ingredients necessary for life--in the powder sample drilled from the "Sheepbed" mudstone in Yellowknife Bay.

Explanation:

Hope it helped

8 0

3 years ago

Read 2 more answers

A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled

Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring (T) = ma

Tension force in the spring (T) = 9 × 2.10

Tension force in the spring (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0

3 years ago

The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the

Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0

3 years ago

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90

babymother [125]

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is $1.56\frac{rev}{s}$

Explanation:

Initial moment of inertia when arms and legs in is $I_i=0.90 kg.m^{2}$

Final moment of inertia when her arms and on leg open outward, $I_f=3.0 kg.m^{2}$

Initial angular speed $w_i=5.2\frac{rev}{s}$

Let the final angular speed be $w_f$

Since external torque on her is zero so we can apply conservation of angular momentum

$\therefore L_f=L_i$

=> $I_fw_f=I_iw_i$

=> $w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}$

Thus her angular speed (in rev/s) when her arms and one leg open outward is $1.56\frac{rev}{s}$

7 0

3 years ago