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3 years ago

14

The go kart driver does 1234J of work when hitting a barrier. Assuming the barrier moved 100 meters from its original location,

what was the force at which the driver hit the barrier?

2 answers:

leva [86]3 years ago

6 0

Force=1234J*100m
So the answer is 123400

bezimeni [28]3 years ago

4 0

The guy in front of me is correct

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Which statement best describes a characteristic of gases?

salantis [7]

Assumes the shape and volume of its container
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3 years ago

A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between

mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

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3 years ago

Explain why does the bowling ball decelerate as it travels along the lane

frutty [35]

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3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t

andre [41]

Answer:

2.5 s, 5 m

Explanation:

The equations for the horizontal and vertical position of Lukalu are:

$x(t) = 8t\\y(t) = -16t^2 + 100$

we can find the time it takes her to reach the ground by requiring that the vertical position becomes zero:

y(t) = 0

So we find:

$0=-16t^2 +100\\16t^2 = 100\\t=\sqrt{\frac{100}{16}}=2.5 s$

The horizontal distance of Lukalu instead will be given by the equation for the horizontal position, substituting t = 2.5 s:

$x=8t = 8 \cdot 2.5 s =5 m$

4 0

3 years ago