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4 years ago

The North Star is located in the Big Dipper. True or False

Physics

2 answers:

harina [27]4 years ago

4 0

Answer:

False

Explanation:

It is located in the little dipper whose stars are more faint.

Maurinko [17]4 years ago

4 0

False, the North Star is located at the end of the handle of the Little Dipper, as opposed to the Big Dipper. Hope this helps!

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An automobile slows from 26 m/s to 18 m/s. Its acceleration is −2.0 m/s 2 . How much time does it take for the vehicle to slow d

Serggg [28]

Answer:4.0s

Explanation:jus is

4 0

3 years ago

The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

3 years ago

If you push a chair across the floor at a cibstant velocity how does the force of friction compare with the force you exert?

Roman55 [17]

-- "constant velocity"   ===> acceleration is 0 .

-- acceleration=0   ===>   forces are balanced

-- balanced forces   ===> (friction) + (force you exert) = 0

   (friction) = -(force you exert)

They have equal magnitude and opposite direction.

7 0

4 years ago

A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he sli

olga_2 [115]

Answer:

His acceleration is $\overrightarrow{a}=4\frac{m}{s^{2}}$

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

$\sum\overrightarrow{F}=m\overrightarrow{a}$

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So $\sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f}$ so (1) is:

$\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}$

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

$W-f=m\overrightarrow{a}$ , solving for a:

$\overrightarrow{a}=\frac{W-f}{m}$ , weight is mg:

$\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}$

$\overrightarrow{a}=4\frac{m}{s^{2}}$

6 0

4 years ago

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.

pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:

R_a=(4*p*L_a)/π*d^2_A (1)

the area of the second wire is:

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i) (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0

3 years ago

Read 2 more answers