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3 years ago

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a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

ed A at the end of this 4.0-second interval? (with work, all steps))

1 answer:

Vika [28.1K]3 years ago

5 0

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

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A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}$

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.

3 0

3 years ago

The sharing of valence electrons between atoms indicates an ionic bond.<br> True or false

morpeh [17]

Answer:

False

Explanation:

Sharing valence electrons to make a bond creates a covalent bond, not an ionic bond.

7 0

3 years ago

Plz help it’s passed due!

pickupchik [31]

It is b 9.1 :) have a nice day

5 0

4 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

3 years ago

Lillian (mass 40.0 kg) standing at rest on slippery ice catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Wha

mestny [16]

Answer:

0.82 m/s

Explanation:

From the law of conservation of momentum,

Total momentum = Final momentum.

mu+m'u' = V(m+m')......................... equation 1

Where m = mass of Lillian, u = initial velocity of Lillian before she catches the dog, m' = mass of the dog, u' = initial velocity of the dog, V = velocity of Lillian and the dog.

make V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Given: m = 40 kg, m' = 15 kg, u = 0 m/s (Lillian was standing at rest), u' = 3.0 m/s.

Substitute into equation 2

V = (40×0+15×3)/(40+15)

V = (0+45)/55

V = 45/55

V = 0.82 m/s

3 0

4 years ago