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rusak2 [61]
3 years ago
13

The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will b

e_______, and, consequently, the control will be better.
The acceleration along the ramp, therefore, the speed at the bottom of a slope will be:__________
a. lower
b. gcostheta
c. higher
d. gsintheta
Physics
1 answer:
pogonyaev3 years ago
3 0

Answer:

Lower

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

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8 0
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The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh
jeyben [28]

Explanation:

Given that,

Electrostatic force, F=3.7\times 10^{-9}\ N

Distance, r=5\ A=5\times 10^{-10}\ m

(a) F=\dfrac{kq^2}{r^2}, q is the charge on the ion              

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}      

q=3.2\times 10^{-19}\ C

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

n=\dfrac{q}{e}

n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}

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Hence, this is the required solution.                        

8 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together
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Answer:

Final velocity = 7.677 m/s

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KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

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u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0
3 years ago
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Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

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