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Simora [160]
3 years ago
5

Select all the correct ANSWERS.

Physics
1 answer:
MissTica3 years ago
3 0
The ground exerts an equal force on the golf ball.
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Y=1/2 at^2 solve for t
Aloiza [94]

Explanation:

y = ½ at²

Multiply both sides by 2:

2y = at²

Divide both sides by a:

t² = 2y/a

Take the square root of both sides:

t = √(2y/a)

6 0
3 years ago
Pls help i will give u brianliest!
Fofino [41]

Answer:

Explanation:

neautrons and protons

3 0
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Crystalline salts such as cesium bromide form _____ bonds with _______.
olasank [31]
Ionic bonds with electrostatic attractions
4 0
3 years ago
Read 2 more answers
Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How
GREYUIT [131]

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed  of 39.2 m/s?

6 0
3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
3 years ago
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