A baseball accelerates downward at 9.8m/s. if the gravitational force acting on the baseball is 2.2n what is the baseballs mass

Pls help me I don’t get it :(

Ksju [112]

Answer:

2nd and 4th

Explanation:

4 0

3 years ago

Why doesn't the skateboarder rolls as high up the ramp each time you go down the ramp

Alexandra [31]

Because of the law of inertia and it’s effect on the skater

5 0

3 years ago

A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass = $volume\times density$

Using the formula

Mass of solution= $1.12\times 473=529.76 g$

Mass % of KBr= $\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100$

Mass % of KBr= $\frac{49.3}{529.76}\times 100$

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

7 0

3 years ago

Identify the chemical change below

lesya [120]

Imma say go with C (:

5 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago