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2 years ago

5

a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti

me interval?

1 answer:

Pepsi [2]2 years ago

7 0

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

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Where should clouds form and why

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Answer:

When air rises in the atmosphere it gets cooler and is under less pressure. When air cools, it's not able to hold all of the water vapor it once was. Air also can't hold as much water when air pressure drops. The vapor becomes small water droplets or ice crystals and a cloud is formed.

Explanation:

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2 years ago

Lithium is more active than aluminium<br> A.True<br> B.false

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3 years ago

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

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3 years ago

the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

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3 years ago

Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in

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We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

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3 years ago

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