Một người đi xe đạp trong 40 phút với vận tốc là 12km/h. Hỏi quãng đường đi được là bao nhiêu?

A person is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the

dangina [55]

here we will use the torque balance about the knee joint

here we can say that

$\tau_g = \tau_m$

here torque due to weight is given as

$\tau_g = 40.1 cos\theta*(l_1 + l_2)$

$\tau_g = 40.1 cos27.5*(0.105 + 0.150)$

$\tau_g = 9.07 Nm$

now torque due to applied force of muscle

$\tau_m = M*sin\alpha * l_1$

$\tau_m = M*sin30* 0.105$

now by torque balance we will have

$9.07 = M*0.5*0.105$

$M = 173 N$

so here the magnitude of m will be 173 N

7 0

3 years ago

An investigation involves determining which metal is better for making pots that will cook food faster. Which is the best hypoth

irina [24]

Because the specific metals aren’t mentioned in this inquiry. The educational guesses that we can propose is that:
1. The hypothetical inquiry: There are existing metals for making pots that will cook food much faster.

2. The one-tailed alternative hypothesis: There are other metals for making pots that will cook food much faster than the other metals. 
3. The one-tailed null hypothesis: All metals that are used in making pots will cook food at an equal rate.

8 0

3 years ago

A 66-foot-tall woman walks at 55 ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

I am Lyosha [343]

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$

5 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

Help me <3 please Thank you :)

NARA [144]

Answer:

11,890

Explanation:

First we need to know what is considered a significant figure.

A significant figure is a value that is not a zero at the start OR end of a value.

Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.

The 0 in the value of 3056 is considered a significant figure.

So from the table, we can deduce:

0.275 has 3 significant figures

750 has 2 significant figures

$10.4 \times {10}^{5} = 1040000$

has 3 significant figures.

11,890 has 4 significant figures.

320,050 has 5 significant figures.

So from the above, we can already see the answer.

4 0

2 years ago