Ask question

Ask question

All categories

3241004551 [841]

4 years ago

14

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.460 m/s. The to

tal mass of the sled, man, and rock is 95.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 15.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving).

2 answers:

xz_007 [3.2K]4 years ago

6 0

Answer:

= 0.417 m/s

Explanation:

Momentum before throwing the rock: m*V = 95.0 kg * 0.460 m/s

= 44.27 N*s

A) man throws the rock forward

mass of rock m1 = 0.310 kg

V1 = 15.5 m/s, in the same direction of the sled with the man

sled and man:

m2 = 95 kg - 0.310 kg = 94.69 kg

v2 = ?

Conservation of momentum:

momentum before throw = momentum after throw

44.27N*s = 0.310kg * 15.5m/s + 94.69kg*v2

⇒ v2 = [44.27 N*s - 0.310 * 15.5N*s ] / 94.69 kg

= 0.417 m/s

AleksandrR [38]4 years ago

6 0

Answer:

0.41 m/s.

Explanation:

Mt = 95 kg

Mr = 0.31 kg

Vr = 15.5m/s

Ut = 0.46 m/s

Mass of the man and sled = (95 - 0.31) kg

= 94.69 kg

Using conservation of momentum equation,

Momentum before the throw = momentum after the throw

95 × 0.46 = 0.31 × 15.5 + 94.69 × V2

43.7 = 4.805 + 94.69 V2

V2 = 0.41 m/s.

You might be interested in

When an object moves, stops moving, changes speed, or changes direction, how do scientists describe that condition?

lorasvet [3.4K]

Drop "moves" from the list for a moment.

You can also drop "stops moving", because that's included in "changes speed"
(from something to zero).

When an object changes speed or changes direction, that's called "acceleration".

I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
there's no acceleration.

I think you meant to say "starts moving". That's a change of speed (from zero
to something), so it's also acceleration.

8 0

3 years ago

Describe how one plays Dr.Dogeball

JulsSmile [24]

•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D

4 0

3 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

4 years ago

A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?

Serhud [2]

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

5 0

3 years ago

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

3 0

3 years ago