All categories

2 years ago

If a car is traveling at a velocity of 5 m/s north , how far wil it travel in 10 seconds?

Physics

2 answers:

11Alexandr11 [23.1K]2 years ago

8 0

Here is your answer

Velocity of car= 5 m/s

Time= 10 secs

We know that

Displacement = Velocity × time

So,

d= v×t= 5×10 m= 50 m

HOPE IT IS USEFUL

Galina-37 [17]2 years ago

5 0

5 m/S ✖ 10 S gives the distance which is 50 m

$5 \times 10$

You might be interested in

How many kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km/h ? 1 kcal = 4

timofeeve [1]

<span>Answer: 1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use. Given: m= 1200kg v initial = 95km/hr v final = 0 2nd, focus on the units - in most cases units speak for the concept the unit of the unknown is kcal, thus its the unit of energy or work so, W = ? 3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present since W = delta K.E =delta P.E W= 0.5m( vf^2 - vi^2) ---> best formula 4th, Substitute the given to the formula since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J we express first 95 km/hr to m/s 95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec W= 0.5(1200kg)[(0^2- (26.39m/sec)^2] W=600 kg(0 - 696.43m^2/s^2) W=600kg(-696.43m^2/s^2) W=417859.3Nm or 417859.3 J W = 417859.3 J x 1 cal /4.19 J W = 99,727.7 cal or 99.728 kcal</span>

4 0

3 years ago

Which was the first object made by humans to orbit earth?

Romashka-Z-Leto [24]

The first human made object made to orbit earth was the Soviet Union's Sputnik 1, it launched the 4th of October 1957.

6 0

3 years ago

A baseball is batted. It's a long fly ball. 4 seconds later the ball reaches the outfield 100 meters away and returns to the hei

Vitek1552 [10]

Answer:

25 m/s

Explanation:

First we should define the variables

T=4

Dx = 100

ay=-9.8

ax=0

We can use formula 1 from the BIG 5

x=(v+v0)t/2

By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

v=v0 since horizontal acceleration always equals zero

so 2v0 = 50

v0 = 25

8 0

2 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$