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2 years ago

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Bears eat fruits such as berries and animals such as fish. They hibernate in the winter. They give birth to live young . Which o

f these terms apply to bears?

1 answer:

bulgar [2K]2 years ago

7 0

Answer:

Bears are consumers

Explanation:

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Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

5 0

2 years ago

Read 2 more answers

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

2 years ago

A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin

aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample $V_1=90\ mL$

Temperature $T_1=298\ K$

Pressure $P_1=702\ mm\ of\ Hg$

for different conditions

Temperature $T_2=273\ K$

Pressure $P_2=760\ mm\ of\ Hg$

suppose $V_2$ is the volume of sample

Using ideal gas equation

$PV=nRT$

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\frac{702\times 90}{298}=\frac{760\times V_2}{273}$

$V_2=76.15\ mL$

8 0

2 years ago

A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is

Radda [10]

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

8 0

1 year ago

A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit

Artyom0805 [142]

Answer: The angle between the wire segment and the magnetic field 66.42°

Explanation:

Please see the attachment below

8 0

2 years ago

Read 2 more answers