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Westkost [7]
4 years ago
9

If a fox starts from rest and accelerates at a rate of 1.1 m/s^2, how long does it take the fox to run 5 m A. 3.0s B. 4.5s C. 9.

1s D) .22s
Physics
2 answers:
KATRIN_1 [288]4 years ago
8 0
Given:
u=0 m/s
a=1.1 m/s^2
S=5 m
t=time it takes to run 5 m

Use the kinematics equation
S=ut+(1/2)at^2
=>
5=0*t+(1/2)1.1(t^2)
solve for t
t=sqrt(5*2/1.1)=3.015 seconds.

kherson [118]4 years ago
3 0

Answer:

3.0 s

Explanation:

Apex

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Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30
vekshin1

66.7 Watts

Explanation:

Let R_{1}=1.0 ohms, R_{2}=3.0ohms and R_{3}=6.0\:ohms. Since R_{2} and R_{3} are in parallel, their combined resistance R_{23} is given by

\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}

or

R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms

The total current flowing through the circuit <em>I</em> is given

I=\dfrac{V_{s}}{R_{Total}}

where

R_{Total}=R_{1}+R_{23}= 3.0\:ohms

Therefore, the total current through the circuit is

I=\dfrac{30\:V}{3.0\:ohms}=10\:A

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor V_{1}:

V_{1}=(10\:A)(1.0\:ohms)=10\:V

This means that voltage drop across the 6-ohm resistor V_{3} is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W

4 0
3 years ago
A dog is walking back home and its motion is modeled by the graph above. What position was the dog at 3 seconds after it started
hammer [34]
6 meters because 6/3 = 10/5
4 0
3 years ago
A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

3 0
4 years ago
Read 2 more answers
The moon’s gravity causes the oceans to bulge. how does this affect tides?
Katarina [22]
The moon affects the tides by causing high and low tides.
3 0
3 years ago
A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No
Mademuasel [1]

Answer:

v_1=-2.616\ m/s

Explanation:

Given that,

Mass of ball 1, m_1=0.25\ kg

Initial speed of ball 1, u_1=5\ m/s

Mass of ball 2, m_2=0.8\ kg

Initial speed of ball 2, u_2=0 (at rest)

After the collision,

Final speed of ball 2, v_2=2.38\ m/s

Let v_2 is the final speed of ball 1.

Initial momentum of the system is :

p_i=m_1u_1+m_2u_2

p_i=0.25\times 5+0

p_i=1.25\ m/s

Final momentum of the system is :

p_f=m_1v_1+m_2v_2

p_f=0.25\times v_1+0.8\times 2.38

p_f=0.25 v_1+1.904

According the law of conservation of linear momentum :

initial momentum = final momentum

1.25=0.25 v_1+1.904

v_1=-2.616\ m/s

So, the final velocity of ball 1 is (-2.616)m/s.

8 0
3 years ago
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