One way we can care for our ear

4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con

Inessa [10]

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km 1000 m 1 hour 1 min

----------- x ------------- x -------------- x ---------- = 44.4 m/s

1 hour 1 km 60 min 60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m

---------- X x sec = 700 m

1 sec

Solve for x sec

x sec = 700m / 44.4 m/s

= 15.8 seconds

3 0

3 years ago

the engine of a car of amass of 2000 kg produced a force of 15000N find the acceleration of the car

weqwewe [10]

I believe the answer would be 7.5 m/s^2

3 0

3 years ago

A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent

Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0

3 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

<u>Given:</u>

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

3 years ago

The Hoover dam is a hydroelectric power plant that converts the energy of falling water into electricity. Which of the following

vlabodo [156]

The correct answer to the question is : B) The weight of the water, and C) The height of the water.

EXPLANATION :

Before coming into any conclusion, first we have to understand potential energy of a body.

The potential energy of a body due to its position from ground is known as gravitational potential energy.

The gravitational potential energy is calculated as -

Potential energy P.E = mgh

Here, m is the mass of the body, and g is the acceleration due to gravity.

h stands for the height of the body from the ground.

We know that weight of a body is equal to the product of mass with acceleration due to gravity.

Hence, weight W = mg

Hence, potential energy is written as P.E = weight × height.

Hence, potential energy depends on the weight and height of the water.

3 0

3 years ago

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One way we can care for our ear​

One way we can care for our ear