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2 years ago

12

A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

s the change in momentum of the car?

1 answer:

antiseptic1488 [7]2 years ago

4 0

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

= 1000 (0 - 20)

= 1000 (-20)

= -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

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Participating in team sports can have a positive effect on all aspects of life. Please select the best answer from the choices p

AveGali [126]

Answer: True

Explanation:

Participating in team sports can lead to the development of healthy fellowship and brotherhood behavior among the participants. It induces team spirit and allows to lead a healthy competition. It can be a good recreational activity which can be used for removing the emotional and physiological stress. Hence, can improve positivity in all aspects of life.

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2 years ago

A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

vazorg [7]

Answer:

8z

Explanation:

It is 8z

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3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 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

2 years ago

A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0

3 years ago