All categories

3 years ago

Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración

Physics

1 answer:

guajiro [1.7K]3 years ago

6 0

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

You might be interested in

A 0.8 m length of wire is formed into a single turn, square loop in which there is a current of 12 A. The loop is placed in magn

rewona [7]

Answer:

0.073 N-m

Explanation:

i = 12 A, l = 0.8 m, B = 0.12 T

The circumference of the loop is 0.8 m.

Let r be the radius of the loop.

2 x 3.14 x r = 0.8

r = 0.127 m

Maximum Torque = i x A x B

Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m

8 0

3 years ago

What is the acceleration of a 3 kg rock that is thrown with a force of 18 N?

Ket [755]

Answer:6m/s^2

Explanation:

Mass =3kg

Force=18N

Acceleration =force ➗ mass

Acceleration =18 ➗ 3

Acceleration =6

Acceleration =6m/s^2

7 0

3 years ago

What type of observation is made through interviewing people’s

Marianna [84]

Answer:

Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.

8 0

3 years ago

Which one of the following is insoluble in water?

Maksim231197 [3]

Answer:

Lemon slices is the correct answer

7 0

3 years ago

Read 2 more answers

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loa

Alex787 [66]

Answer:

(a)

W_friction = 98.1 J

W_net = 550.9 J

(b)

W_friction = 98.1 J

W_net = 463.95 J

Explanation:

(a)

First, we will calculate the work done by friction:

$W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\$

W_friction = 98.1 J

Now, the work done by Eskimo will be:

$W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\$

W_Eskimo = 649 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 649 J - 98.1 J

W_net = 550.9 J

(b)

First, we will calculate the work done by friction:

$W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\$

W_friction = 98.1 J

Now, the work done by Eskimo will be:

$W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\$

W_Eskimo = 562.05 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 562.05 J - 98.1 J

W_net = 463.95 J

4 0

3 years ago