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Dahasolnce [82]
3 years ago
15

A 66.0−kg short-track ice skater is racing at a speed of 10.0 m/s when he falls down and slides across the ice into a padded wal

l that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?
Physics
1 answer:
dexar [7]3 years ago
7 0

Answer:

3300J

Explanation:

Work done is the energy that is lost by the skater

Formula for workdone = 1/2*mV^2

m = 66kg

V = 10m/s

Work done = 1/2 * 66 * 10^2

= 3300J

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A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe
Vesna [10]

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

7 0
3 years ago
The density of water is 1.00 g/cm3. What is its density in kg/m3?
r-ruslan [8.4K]
1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.
3 0
3 years ago
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N

So, the force has a magnitude of 4212 N

4 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that
irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0
3 years ago
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