If object a has more mass than object to be does object contain more matter explain

The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.

wolverine [178]

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

$\rm 1 \: g = {10}^{ - 3} \: kg \\ \\ \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6 \times {10}^{ - 3} \ kg \\ \\ \rm = 1.440 \times 10^{6 - 3} \ kg \\ \\ \rm = 1.440 \times 10^3 \ kg \\ \\ \rm = 1.440 \times 1000 \ kg \\ \\ \rm = 1440 \ kg$

5 0

2 years ago

Two students walk in the same direction along a straight path at a constant speed. One walks at a speed of 0.90 m/s and the othe

mr Goodwill [35]

Answer:

A. 456 seconds

Explanation:

We are given that two students walk in the same direction along a straight path at a constant speed.

One student walks with a speed=0.90 m/s

second student walks with speed=1.9 m/s

Total distance covered by each students=780 meter

We have to find who is faster and how much time extra taken by slower student than the faster student.

Time taken by one student who travel with speed 0.90 m/s= $\frac{780}{0.90}$

Time= $\frac{distance}{speed}$

Time taken by one student who travel with speed 0.90 m/s

= $\frac{780}{0.90}$

Time taken by one student who travel with speed 0.90 m/s

=866.6 seconds

Time taken by second student who travel with speed 1.9 m/s= $\frac{780}{1.9}$

=410.5 seconds

The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .

Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds

Extra time taken by the student travels with speed 0.90 m/s=456 seconds

Hence, option A is true.

7 0

2 years ago

Read 2 more answers

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

2 years ago

two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist

Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

$R_{eq}=R_1+R_2$

For parallel combination,

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

When 6 ohm and 3 ohm are in series,

$R_s=6+3\\\\R_s=9\ \Omega$

When 6 ohm and 3 ohm are in paralle,

$\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega$

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0

2 years ago

A coin is thrown with a velocity of 0 m/s down a dry well and hits bottom in 1.2s, what’s the depth of the well?

pentagon [3]

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

$d = \frac{1}{2}at^2+v_0t+d_0$

where d stands for distance, t for time, a for acceleration, v0 and d0 for initial velocity and distance, respectively. Since the initial distance and velocity are both zero, we are left with the first term. The coin is in free fall and so it is accelerated by gravity:

$d = \frac{1}{2}at^2= \frac{1}{2}gt^2=\frac{1}{2}9.8\frac{m}{s^2}1.2^2s^2=7.1m$

The well is 7.1 meters deep.

5 0

3 years ago