Ask question

Ask question

All categories

3 years ago

6

A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?

1 answer:

solmaris [256]3 years ago

5 0

B. 1520 is the difference between their weights.

You might be interested in

How tightly does mass need to be compacted in order to become a black hole??? (2 words)

olga55 [171]

is it infinite density?

8 0

3 years ago

A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago

In a chemical reaction, molecules of hydrogen gas (H2) react with molecules of oxygen gas (O2) in a sealed reaction chamber to p

11Alexandr11 [23.1K]

Answer:

Option (B) is correct.

Explanation:

Given that the molecules of hydrogen gas ( $H_2$ ) react with molecules of oxygen gas ( $O_2$ ) in a sealed reaction chamber to produce water ( $H_2O$ ).

The governing equation for the reaction is

$2H_2 +O_2 \rightarrow 2H_2O$

From the given, the only fact that can be observed that 2 moles of $H_2$ and 1 mole of $O_2$ reacts to produce 2 moles of $H_2O$ .

As the mass of 1 mole of $H_2 = 2$ grams ... (i)

The mass of 1 mole of $O_2 = 32$ grams ...(ii)

The mass of 1 mole of $H_2O = 18$ grams (iii)

Now, the mass of the reactant = Mass of 2 moles of $H_2$ + mass 1 mole of $O_2$

$= 2 \times 2 + 32$ [ using equations (i) and (ii)]

$=4+32 = 36$ grams.

Mass of the product = Mass of 2 moles of $H_2O$

$=2\times 18$ [ using equations (iii)]

=36 grams

As the mass of reactants = mass of the product.

So, mass is conserved.

Hence, option (B) is correct.

8 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

Why do we learn useless nonsense school, why don't we learn what we will actually use in life

laila [671]

I totally agree but, in my opinion its because of the government and what the state has control over. Teacher have little control over it.

4 0

3 years ago