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Marrrta [24]
3 years ago
14

Max (15 kg) and Maya (12 kg) are ice-skating on a frozen pond. While standing at the center of the pond, Maya throws a 1.5-kg sn

owball at Max and, as a result, recoils away from Max at 2.5 m/s. With what speed did Maya throw the snowball at Max
Physics
1 answer:
pychu [463]3 years ago
4 0

Answer: The speed at which Maya threw the snowball at Max is 20m/s

Explanation:

Given that

​Mass of snowball=   1.5-kg.

Mass of Maya= 12kg

Initial  speed of Maya=2.5m/s

We use the law of conservation of  momentum to calculate the  speed of the snowball which depends on its own mass,  the mass of Maya who threw the snowball and the initial velocity at which the snowball was thrown by Maya

Ms x Us = Mm x Um

Mass of snowball  x Initial speed of Snow ball =Mass of Maya x Initial  speed of Maya

1.5kg  x Initial speed of Snow ball =12 kg x 2.5m/s

Initial speed of Snow ball=( 12 kg x 2.5m/s) /1.5kg

​ Initial speed of Snow ball  =20m/s

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Question 4 is true, question 5 is B.
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3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
A wire with mass 50 g is stretched so that it’s ends are tied down at points 100 cm apart. The wire vibrates in its fundamental
ivolga24 [154]
I think the answer is 2500 N
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3 years ago
Read 2 more answers
What would be the daughter product of the beta decay of lead-212 and what would be the daughter product of the alpha decay of po
kati45 [8]

Answer:

lead with Z = 82 is transformed into Bismuth with Z = 73

The Po with Z = 84 becomes Pb with Z = 82

Explanation:

Beta decay occurs when a neutron emits an electron and an anti neutrino from the atomic nucleus, therefore the atomic number of the material increases by one unit.

           Pb + e + ν → Bi

lead with Z = 82 is transformed into Bismuth with Z = 73

In alpha decay, a helium nucleus is emitted from the nucleus of the atom, therefore the atomic number decreases by two units.

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4 0
3 years ago
A man pushes an 250-N crate at constant speed a distance of 30.0 m upward along a rough slope that makes an angle of 60° with th
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Answer:

6495.19 Joule

Explanation:

F = Weight of the crate = 250 N

d = Distance the cart is pushed = 30 m

θ = Angle of inclination = 60°

The weight of the crate will be resloved into two components

Fdsinθ and Fdcosθ

Work done by the force of gravity is

W = Fdsinθ

⇒W = 250×30×sin60

⇒W = 6495.19 Joule

∴ The work done by the force of gravity is 6495.19 Joule

8 0
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