Ask question

Ask question

All categories

3 years ago

5

where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distan

ce between them, and G is the gravitational constant. In SI units, the units of force are kg⋅m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?

1 answer:

stepan [7]3 years ago

4 0

In SI units, the universal gravitational constant is

6.67 x 10⁻¹¹ newton-meter² / kilogram² .

The unit can also be written as meter³ / kilogram-second² .

You might be interested in

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0

3 years ago

A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

RSB [31]

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

5 0

3 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

4 0

3 years ago

Please Please Help!!, What happens when a current moves through a wire and the wire is placed near a magnet?

USPshnik [31]

I think it is c not sure

7 0

3 years ago

Read 2 more answers

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

$Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)$

$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

3 0

3 years ago