Graphs are ____ of relationships.

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The

uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force $F_{net} =\sum F = F_1 +F_2 +..... = ma$

therefore we have from above equation $F_{net} = F -f = ma$

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

8 0

3 years ago

Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

RideAnS [48]

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 12.6 N

d = 15.4 m

$\theta = 15.3 degree$

so we will have

$W = 12.6 \times 15.4 cos15.3$

$W = 187.2 J$

5 0

3 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago

The Celsius temperature scale is based on which of the following

Lapatulllka [165]

Freezing.................

6 0

3 years ago

A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,

AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

$Q_{water}=Q_{methanol}$

where:

$Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})$

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

$Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})$

Now replacing:

$0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]$

4 0

2 years ago