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2 years ago

6

What sport was inspired by basketball, and adapted for older adults who were still interested in being physically active?

1 answer:

Andrew [12]2 years ago

5 0

Answer:

A. Volleyball.

Explanation:

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If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

maks197457 [2]

"decreasing the distance of the space shuttle from Earth"

F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases.

5 0

3 years ago

Read 2 more answers

A girl throws a tennis ball upward with an initial velocity of 4 m/s. What is the maximum displacement of the ball?

Jobisdone [24]

Answer:

0.82 m

Explanation:

The ball is in free fall - uniform accelerated motion with constant acceleration downward, $a=g=-9.8 m/s^2$ (acceleration of gravity). So we can use the following suvat equation to solve the problem:

$v^2-u^2=2as$

where

v is the final velocity

u = 4 m/s is the initial velocity

a is the acceleration

s is the displacement

At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:

$s=-\frac{u^2}{2a}=-\frac{4^2}{2(-9.8)}=0.82 m$

8 0

2 years ago

What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?

pentagon [3]

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In 100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution $\rho _s=1.115\times density\ of\ water$

Thus

$\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL$

mass of sample $M=1.115\times 150=167.25\ gm$

mass of sucrose $m_s=0.35\times 167.25=58.53\ gm$

mass of Water $m_w=108.71 gm$

Volume of water $=108.71\times 1=108.781 mL$

7 0

2 years ago

What happens when you break a magnet in half

Vlad1618 [11]

They'll still be magnets, but they'll never be able to touch each other where they were cut.

I hope this helps you! :-)

6 0

2 years ago

Read 2 more answers

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago