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Alona [7]
3 years ago
5

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

Physics
1 answer:
ioda3 years ago
3 0
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
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Answer:

Land heats faster

Explanation:

Land heats faster because absorbs much more solar radiation.

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All 2023 ariya ac synchronous drive motors produce ____% torque at 0 mph for impressive off-the-line acceleration and smooth cru
Savatey [412]

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<h3>How to calculate the torque?</h3>

Mathematically, the torque of an automobile vehicle can be calculated by using this formula:

Torque = Fd

<u>Where:</u>

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8 0
2 years ago
A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the
ivolga24 [154]

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

4 0
3 years ago
A convex lens with a focal length of 10 cm is located 30 cm away from a
zloy xaker [14]

Answer: 15 cm

Explanation:

According to the Lens Equation we have the following:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u} (1)

Where:

f=10 cm is the focal length

u=30 cm is the distance between the candle (the object) and the lens

v is the distance between the image and the lens

Isolating v:

v=\frac{uf}{u-f}} (2)

Solving:

v=\frac{(30 cm)(10 cm)}{30 cm-10 cm}} (3)

Finally:

v=15 cm This is where the image is located

8 0
2 years ago
You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car
gogolik [260]

B. The sound of the engine will get louder and the pitch higher.

4 0
3 years ago
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