Ask question

Ask question

All categories

melamori03 [73]

3 years ago

8

Basic Question please help

1 answer:

Dvinal [7]3 years ago

7 0

1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)

You might be interested in

PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer

4vir4ik [10]

Answer:

b

Explanation:

3 0

3 years ago

Read 2 more answers

A renewable item is something that is capable of being replaced naturally.

djverab [1.8K]

The answer is False.

6 0

4 years ago

Read 2 more answers

Dial CALIPER ZOOM INNNN

taurus [48]

Answer:

1.142 and 5.621

.

Explanation:

........

8 0

3 years ago

Define the work Envelope for a Robot

Masteriza [31]

A robot's work envelope is its range of movement. It is the shape created when a manipulator reaches forward, backward, up and down. These distances are determined by the length of a robot's arm and the design of its axes. ... A robot can only perform within the confines of this work envelope.

7 0

3 years ago

Please find the power (in kW) needed in accelerating a 1000 kg car from 0 to 100 km/hour in 10 seconds on a 5% gradient up-hill

hodyreva [135]

Answer:

P= 45.384 kW

Explanation:

given data:

m = 1000 kg

u = 0

v = 100 km/hr = 250/9 m/s

t = 10 sec

g =9.81 m/s2

5% gradient

from figure we can have

$tan\theta = \frac{0.05x}{x}$

$\theta = tan^{-1}0.05$

$\theta = 2.86$ from equation of motion we have

v = u + at

$\frac{250}{9} = 0 + 9*10$

$a = \frac{25}{9} m/s^2$

distance covered in 10 sec

from equation of motion

$s = ut + \frac{1}{2}at^2$

$s = 0*10+ \frac{1}{2}*\frac{25}{9}*10^2$

s = 138.8 m

from newton's 2nd law of motion along inclined position

$F -mgsin\theta = ma$

solving for f

$f = mgsin\theta +ma$

$F = 1000*9.81*SIN2.8624 +1000*\frac{25}{9}$

F = 3267.67 N

work done is given as W

$W = F* s$

and power $P = \frac{W}{t}$

$P = \frac{F*s}{t}$

$P = \frac{3267.67*\frac{1250}{9}}{10}$

P = 45384.30 W

P= 45.384 kW

4 0

4 years ago