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Sonbull [250]
3 years ago
13

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
Physics
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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A parallel-plate capacitor is formed from two 2.7 cm -diameter electrodes spaced 1.4 mm apart. The electric field strength insid
Andre45 [30]

Answer:

The potential difference between the plates is 8.4\times10^{3}\ V

Explanation:

Given that,

Distance = 1.4 mm

Electric field strength E= 6.0\times10^{6}\ N/C

Let the potential difference is V.

We need to calculate the potential difference between the plates

Using formula of electric field

E=\dfrac{V}{d}

V=Ed

Where, V = potential

d = distance

Put the value into the formula

V=6.0\times10^{6}\times1.4\times10^{-3}

V=8.4\times10^{3}\ V

Hence, The potential difference between the plates is 8.4\times10^{3}\ V

3 0
3 years ago
Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches
aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass m=380kg

Height of supporting Rock X=85cm

Length of BoardL_r=4.5m

Mass of board M_b=22kg

Mass of adult M_a=62  

Generally the moment of balance about wedge part about  is mathematically given by

N -Q + R = Mg + mg

0.85*N - Mg*2.25 - mg*(2.25 + x) = 0

0.85*N  = + Mg*2.25 + mg*(2.25 + x)

where

N+R=4547

therefore

N = 570.70588 + 1608.3529 + 714.823 x

if N=0 at fallen person

x=3.04m

Therefore it is save to carry a 62kg adult

8 0
2 years ago
A 1.50 x 109 kg railroad car moving at 7.00 m/s to the north collides with
barxatty [35]

GIVE ME YOUR MONEY

Explanation:

8 0
3 years ago
Need help on this thank you
Semmy [17]

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

7 0
3 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17
Alexus [3.1K]

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

8 0
2 years ago
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