3) If an object has a net negative charge of 4.0 Coulombs, the object possesses

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone

Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, $a_c$ , is given as follows;

$a_c = \dfrac{v^2}{r}$

Therefore, the centripetal acceleration of the stone found as follows;

$a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2$

The centripetal acceleration of the stone, $a_c$ = 5 m/s².

5 0

3 years ago

6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove

Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0

1 year ago

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

6 0

3 years ago

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet

Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × $10^{5}$ Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × $10^{5}$ = 13.6 × 10³ × 9.81 × h

h = 0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × $10^{5}$ Pa

6 0

3 years ago

When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi

vova2212 [387]

Answer:

a)

125.6 rad/s

b)

25.12 rad/s²

Explanation:

a)

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = $1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}$

w = $\frac{1200\times 2\pi }{60}\frac{rad}{s}$

w = 125.6 rad/s

b)

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

5 0

3 years ago