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erik [133]
3 years ago
15

If you walk 8km at an average speed of 4km/h, how much time does it take

Physics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0
Distance = rate x time
8km = 4km/h x time
Time = 2 hours
Galina-37 [17]3 years ago
5 0

Answer:

2 hours

Explanation:

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What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?
Kazeer [188]

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

5 0
2 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
Which has the most momentum?
boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass m is moving at a velocity of v, the momentum of that object would be m\, v.

The 100\; {\rm g} (0.1\; {\rm kg}) object is moving at a speed of 1\; {\rm m\cdot s^{-1}}. The magnitude of the momentum of this object would be 0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

Similarly, the momentum of the 1\; {\rm g} (10^{-3}\; {\rm kg}) object moving at a speed of 100\; {\rm m\cdot s^{-1}} would be 10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}.

Hence, the magnitude of momentum is the same for the two objects.

7 0
1 year ago
a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²​
tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

{v_f}^2-{v_i}^2=2a\Delta y

which tells us the player's initial velocity v_i is

0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}

The player's height at time t is given by

y=v_it-\dfrac g2t^2

so we find their airtime to be

0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s

6 0
3 years ago
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
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