Ask question

Ask question

All categories

Salsk061 [2.6K]

2 years ago

12

If a car is traveling at a constant speed of 100 kilometers per hour (km/h), how many hours will it take for the car to travel a

distance of 500 kilometers?

2 answers:

Elena-2011 [213]2 years ago

8 0

Answer:

5 hrs

Explanation:

distance / rate = time

500 km / 100 km/hr = 5 hours

Ede4ka [16]2 years ago

6 0

Answer:

i believe it is 5

Explanation:

because if it moves at 100 miles per hour after 5 hours you will have 500 miles

because 5 multiplied by 100 equals 500

You might be interested in

In a series circuit with three bulbs,

marysya [2.9K]

<span>In a series circuit with three bulbs, </span><span> </span><span>the remaining two bulbs are not affected if one bulb burns out.</span>
If a bulb goes out in your house, do all other bulbs go out? No.
<span>Final answer: B</span>

3 0

3 years ago

Read 2 more answers

Is the force of gravity stronger on a piece of crumpled paper or a normal piece of flat paper?

GaryK [48]

Yes, because a crumpled piece of paper has more center mass, allowing it to fall faster.

3 0

3 years ago

What’s is the movement of one object around another

allsm [11]

Answer:

revolution

Explanation:

8 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago