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Lapatulllka [165]
2 years ago
15

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

Physics
1 answer:
Talja [164]2 years ago
3 0

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

\frac{10N}{50kg}=a

A newton is just \frac{kg·m}{s^{2}}

a=\frac{\frac{10kg·m}{s^{2}}}{50kg}

The s^2 and 50 kg you multiply

a=\frac{10kg·m}{50kg·s^{2}}

The kg's cancel and 10/50 is 1/5

\frac{1}{5}·\frac{m}{s^{2}}

So the acceleration is 1/5 m/s^2


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(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

h=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
3 years ago
A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti
matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

              \phi  = \frac{Q}{\epsilon_{o}}    (Gauss's Law)

Putting the given values into the above formula as follows.

            Q = \phi \times \epsilon_{o}

                = -953 Nm^{2}/C \times 8.854 \times 10^{-12}

                = -8.4 \times 10^{-9} C

                = -8.4 nC

Therefore, when the unknown charge is q  then,

         -14.0 nC + 33.0 nC + q = -8.4 nC

               q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0
3 years ago
The Southwest Indian Ridge, shown in red, moves at an average rate of 20 mm/year, making it among the ultraslow spreading ridges
blagie [28]

v = average speed of movement of the Southwest Indian Ridge = 20 mm/year

d = distance moved by the Southwest Indian Ridge = 100 mm

t = number of years required to move distance "d"

distance traveled is given as

d = v t

inserting the above values in the formula

100 mm = (20 mm/year) t

dividing both side by 20 mm/year

t = 100 mm/(20 mm/year)

t = 5 years

7 0
2 years ago
Read 2 more answers
Neptune is approximately 4.5 billion kilometers from the sun. What is this distance in AU?
asambeis [7]
Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?

4,500,000,000/149,597,870=30.080642 AU.

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7 0
2 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
3 years ago
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