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rusak2 [61]
3 years ago
8

Please help! I have another part to this too.

Physics
1 answer:
JulsSmile [24]3 years ago
5 0

first one is true, there's no net force acting on it thats greater than another or making it unbalanced, if there was the object would be in some kind of motion

All scientist use meters, that way scientist can share information across country without needing to convert the data.

3. is air resistance

4. The large rock

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This is for physics, i need help fast!!
jarptica [38.1K]
Speed = distance / time
S= 40 000m / 5400s
S= 7.41m/s
7 0
3 years ago
Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m
frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

L = - 1361.591 k Kgm/s

7 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
Please help! I'm not sure what equation or the process to do this question.
lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

5 0
3 years ago
(a) Determine the specific volume of nitrogen gas at 8.5 MPa and 139 K based on the generalized compressibility chart (Fig. A-15
Vikentia [17]

Answer:v=0.0485m^3/kg

Explanation:

PV=MRT

PV/M=RT

Pv=RT

P=8.5mpa= 850Kpa

T= 139K

v= RT/P

0.2968*139/850

v=0.0485m3/kg

Error

0.0485-0.002002/0.0485*100%

95%

4 0
3 years ago
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