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Masja [62]
2 years ago
10

What force can stop a 1400kg truck accelerating at 7.5m/s2? URGENT PLZZ

Physics
2 answers:
Archy [21]2 years ago
6 0

Answer:

10500 N

Explanation:

F = m*a=1400*7.5N=10500N

GrogVix [38]2 years ago
5 0
10500 N I think is your answer
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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground
sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

7 0
3 years ago
Dams are made wider at the bottom than at the top?
malfutka [58]
Yes dams are made wider at the bottom because the pressure of the water pressure is greater there
8 0
3 years ago
You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi
Elenna [48]

Answer:

\frac{g_{2}}{g_{1}} = \frac{1}{4}

Explanation:

The period of the simple pendulum is:

T = 2\pi\cdot \sqrt{\frac{l}{g} }

Where:

l - Cord length, in m.

g - Gravity constant, in \frac{m}{s^{2}}.

Given that the same pendulum is test on each planet, the following relation is formed:

T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}

\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}

\frac{g_{2}}{g_{1}} = \frac{1}{4}

5 0
3 years ago
PLEASE ANSWER ASAP!.!.!!.!!!!.!.!.
lbvjy [14]
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8 0
3 years ago
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