What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Mathematics

2 answers:

Likurg_2 [28]3 years ago

8 0

Force = mass x exceleration
1100 x 0.5 = 550

AlexFokin [52]3 years ago

4 0

Answer: 550 N

Identify the given information.
m = 1100 kg
a = 0.5 m/s²

Choose which formula you should use. In this case, you are finding the force.
f = ma, where m is the mass and a is the acceleration.

Substitute and solve.
f = (1100 kg)(0.5 m/s²)
f = 550 N

You might be interested in

Find the product of 9x(-3x²y-7xz)

kogti [31]

Answer:

$- 27 {x}^{3} y - 63 {x}^{2} z$

Step-by-step explanation:

$9x( - 3 {x}^{2} y - 7xz)$

$= 9x \times ( - 3 {x}^{2} y) + 9x \times ( - 7xz)$

$= - 27 {x}^{3} y - 63 {x}^{2} z$

4 0

3 years ago

In a bowl, there are 3 peaches, 2 apples, 6 lemons, and 4 plums. If 1 piece of fruit is to be selected at random, what is the pr

wolverine [178]

All together there are 15 pieces of fruit
out of 15, 6 of them are lemons.
the probability that the fruit will be a lemon is
6/15
which reduces to
2/5

3 0

3 years ago

The radius of a semicircle is extended by 6 inches.how does the measure of the semicircles arc length change?

vivado [14]

Let x be the original radius. Given that a semicircle will have an arc length of half of a circumference and C=2pr, the arc length will be just pr.

Now the difference of arc is then going to be:

p(6+x)-px

p(6+x-x)

6p inches

So the arc will be 6p inches longer. Or if you wish to approximate...

≈18.85 inches

7 0

3 years ago

A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms

MissTica

Answer:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05